Chapter 15: Electric Field: Force and Energy Approaches

Etkina/Gentile/Van Heuvelen Process Physics 1/e, Chapter 15 15-1

Chapter 15: Electric Field: Force and Energy Approaches

! Why does a metal pole on the top of a building protect it from lightning? ! How does electrocardiography work? ! Why is it safe to sit in a car during a lightning storm? Make sure you know how to: 1. Find the force that one charged object exerts on another charged object. 2. Determine the electric potential energy of a system. 3. Explain the difference between the internal structure of electric conductors and dielectrics (non-

conductors).

Chapter opening Imagine that you are in a car during a thunderstorm. The thunder is loud; you see lightning strikes on the road just in front of you—huge compared to the sparks you see when removing clothes from the dryer. Should you be scared for your life and worried about lightning damage to the fancy electronics in your car? The frame of the car is made of metal, which is a good electrical conductor. We will learn in this chapter why the insides of cars made of conducting material are safe places to be during a thunderstorm, even if lightning strikes the car directly. It turns out that the internal structure of conductors is what allows them to shield what is inside them from all electrostatic processes happening outside.

Lead In Chapter 14 we learned how to describe the electric interaction in two ways: 1) by determining a force exerted by one charged object on another, and 2) by determining an electric potential energy possessed by pairs of charged objects. We learned that electric charge is a property of particles. When we add or remove these particles from macroscopic objects, the objects can then participate in electric interactions. We also learned that the force that these charged objects exert on each other depends on the magnitude of their charges, and also on the distance between them. Similarly, the electric potential energy possessed by pairs of charged objects depends on their charges and the distance between them. We also discovered that charged objects exert forces on each other without being in direct contact. This is only the second interaction we have encountered with this property, the gravitational interaction being the first. How does one charged object “know” about the presence of another when they are not in direct contact? We investigate this question in this chapter.

15.1 Mechanisms for the electric interaction In the previous chapter we learned that the magnitude of the force that two electrically-

charged objects exert on each other is described by Coulomb’s law: 21 on 2 2 on 1 1 2F F kq q r" " .

Assuming that electrically charged objects (such as the electrodes of a Wimshurst generator or a


balloon that you rubbed on your hair) don’t have some sort of supernatural sense that allows them to “see” their surroundings, how is it that they “know” to interact with other charged objects when those other objects could be meters away? Let’s start our investigation into this phenomenon with observational experiments. Observational Experiment Table 15.1 Disrupting electric force.

Observational experiment Analysis (1) Rub a plastic rod with felt (it becomes negatively charged) and bring it close to the top of an electroscope without touching it. The leaves of the electroscope deflect. When you remove the rod, the leaves go back.

The rod exerts an electric force on the movable electrons in the electroscope, pushing them down into the leaves. The negatively charged leaves repel each other. In the absence of the rod, the electrons redistribute back.

(2) Rub a plastic rod with felt and bring it close to the top of an electroscope whose top is covered with (but not touching) a glass cup. The leaves deflect less than in the first experiment. When you remove the rod, the leaves go back.

The rod exerts a smaller electric force on the movable electrons in the electroscope. The bottom parts become less negatively charged and repel each other less. In the absence of the rod, the electrons redistribute back.

(3) Rub a plastic rod with felt and bring it close to the top of an electroscope whose top is covered with (but not touching) a metal pop can. The leaves do not deflect.

When the top of the rod is covered with the metal can, for some reason the rod exerts no electric force on the movable electrons in the electroscope.

Pattern It appears that different materials placed between charged objects affect the intensity of the interaction. Metals have the biggest effect—reducing the interaction to zero.

How can we explain why the metal can covering the top of the electroscope makes it insensitive to the presence of the charged rod outside? (Fig. 15.1a) Here are two possible explanations.


Figure 15.1(a) Metal cup covers top of electroscope

Explanation (model) 1 The metal can (an electrical conductor) itself has movable electrons. The negatively charged rod causes free electrons in the can to move down the can on its outside leaving a deficiency of electrons (positive charge) on the top of the can (Fig. 15.1b) and an excess of electrons on the bottom. The net electric force that both the charged rod and the polarized can exert on the electrons in the electroscope is zero (Fig. 15.1c).

Figure 15.2(b)(c)

Explanation (model) 2 The can shields the top of the electroscope from the influence of the charged rod by “blocking” the interaction. This is similar to how soundproofing can shield a recording studio room from sound coming from outside.

The distinction between these two explanations is that the first involves the idea of interaction without contact but does not address the question of how this happens. The second explanation answers the question how by suggesting that there is something between the charged objects responsible for their interaction and this something can be blocked.

Let’s think about these two explanations (models) more carefully. We know that electrically charged objects interact without directly touching each other. How does one charged object at a certain location “feel” the presence of another charged object located somewhere else? Historically, there have been two answers to this question, corresponding to the two models above. The first model was called interaction at a distance. Sir Isaac Newton (in the case of the gravitational interaction) and Charles Coulomb (in the case of the electric interaction) supported this model. In this model, the


electric interaction just happens. If you move one of the two interacting charges, the other charge instantaneously “senses” that movement and responds accordingly.

Though model 1 doesn’t suggest how this interaction happens, it can explain the disruption of the interaction between charges when a metal can surrounds one of those charges. The freely moving electrons in the metal can rearrange themselves in the presence of the charged plastic rod on the outside (see Fig. 15.1b and c). The electric charges present on the surface of the metal can exert an electric force on the electrons in the electroscope that just balances the force exerted by the charged plastic rod outside; the net electric force exerted on the electrons in the electroscope rod under the can is zero.

The second model for electric interactions involves an agent that acts as an intermediary between the charges; some sort of electric disturbance in the region. Physicists call this electric disturbance an electric field. According to this ‘electric field model’, a charged object Q electrically

disturbs the region surrounding itself (Fig. 15.2a). If you place a second charged object q in this

region, the electric field exerts a force on it (Fig. 15.2b). This model suggests that the metal can in Experiment 3 of Table 15.1 altered the electric field in the region surrounding the top of the electroscope (inside the can) in such a way that it resulted in a zero net electric force being exerted on the electrons inside the electroscope, despite the presence of the charged rod on the outside. By suggesting the existence of the electric field, this ‘electric field model’ gives a mechanism for the electric interaction (as opposed to the ‘interaction at a distance model’ which had no mechanism for magically exerting a force on another distant charged object.)

(a) (b) Figure 15.2 Electric field model for interactions between electric charges

Both models account for the outcomes of the observation experiments (Table 15.1.) This

happens frequently in science; when multiple models can adequately explain all observations made up to that time. In later chapters we will encounter phenomena that are not easily explained with the interaction at a distance model, and are more easily explained with the electric field model. Because of this, we will use the electric field model from now on.

Here’s an analogy to help visualize the idea of a field. Imagine a large horizontal rubber sheet that has been pulled tight. This represents the electric field. Now, place a bowling ball in the center of the sheet. This causes the sheet to bend, and it is bent more severely closer to the bowling ball than further away from it. This represents the way a charged object alters the electric field in the region surrounding it. If you placed a marble at various places on the rubber sheet you would find that when the marble is closer to the bowling ball it accelerates toward it faster than when it is further away.


This is consistent with our understanding of the electric interaction; we have learned that the electric force that charges exert on each other is greater when the charges are closer. Notice how the rubber sheet is acting as an intermediary, allowing the bowling ball and the marble to interact with each other even though they are not in contact. This is analogous to the way the electric field acts as an intermediary between charged objects, allowing them to interact even though they are not in contact with each other.

We have encountered one other interaction in which two objects can interact without being in physical contact: the gravitational interaction. We can use the field model to explain how this is possible. For example, how does Earth “know” to travel in an orbit around the Sun when there is a quarter of a 150 million kilometers (about 90 million miles) between them? The field model suggests that the Sun gravitationally (rather than electrically) disturbs the region around it in an invisible way. This disturbance is called a gravitational field. So, the Sun produces the gravitational field in the region surrounding it (Fig. 15.3a), and the gravitational field exerts a force on Earth causing it to travel in an elliptical orbit (very close to a circle) around the Sun (Fig. 15.3b).

Figure 15.3 Field approach for gravitational force

Each charge produces a field and Newton’s third law

So far, we have focused on the field produced by one charged object and the force that field exerts on a second charged object. Why choose one charged object as the source of a field (and not the other charge) and the other charged object as the recipient of the force due to that field? Consider again two electrically charged objects (1 and 2) that interact with each other (Fig 15.4a.) First let’s choose the system of interest to be object 1. This means object 2 is part of its environment. In terms of the electric field model (model 2), we would say that charged object 2 produces an electric field in the region surrounding itself (Fig. 15.4b); as a result, that electric field exerts a force on object 1 (Fig. 15.4c). In the field model, object 2 is not exerting a force directly on object 1—its field is exerting the force.

Figure 15.4(a)(b)(c) Charges 1 and 2 both produce fields that exert forces on other charges


Let’s now choose the system of interest to be object 2. The electric field model says that object 1 produces another electric field in the region surrounding itself (Fig. 15.4d); as a result, the electric field produced by charge 1 exerts a force on object 2 (Fig. 15.4e). Each charge produces its own electric field and that field exerts a force on other charges in the field. Qualitatively, this is consistent with Newton’s 3rd law: object 1 exerts an electric force on object 2, and object 2 in turn exerts an electric force on object 1.

Figure 15.4(d)(e)

Now, what about an arbitrary point somewhere between objects 1 and 2. Is there an electric

field there? How does it relate to the field produced by 1 and 2? To answer those questions, we need to learn how to characterize electric field mathematically. Review Question 15.1 What is the difference between the two models of the electric interaction: the interaction at a distance model and the electric field model?

15.2 The E!

-field In the previous section we devised a new concept, the electric field as a mediator of

electrostatic interactions and the gravitational field as a mediator of the gravitational interactions. The goal of this section is to construct a mathematical description of an interaction based on the field model. We start with the more familiar gravitational interaction.

Consider Earth and its contribution to the gravitational field. We say ‘contribution’ because the Sun, the Moon, the other planets, etc. all contribute to the gravitational field as well. But, our interest for the moment is in the region near Earth, so Earth’s contribution to the gravitational field will be the dominant one; we assume the contributions from all other objects are minimal.

Imagine Earth and one of the three objects A, B or C near Earth (Fig. 15.5a.) The masses of

the objects are Am m" , B 2m m" , and 3Cm m" .

Figure 15.5 Earth exerts gravitational force on one of three objects placed at same location


Consider the gravitational force that Earth exerts on each object. We are considering Am , Bm , and

Cm individually as our systems of interest. Earth then is part of the environment and its gravitational

field exerts a force on A, B, or C (Fig. 15.5b). Let’s compare these three gravitational forces using the law of universal gravitation:

E EE on A A A2 2

E EE on B B B2 2

E EE on C C C2 2

m mF G m m Gr rm mF G m m Gr rm mF G m m Gr r

# $" " % &' (# $" " % &' (# $" " % &' (

Notice that the magnitudes of these forces are proportional to the masses of the objects on which they are exerted. The directions of these forces are toward the center of Earth and are in exactly the same direction if the three objects are placed one at a time at the same location. However, the magnitudes of the forces exerted on them differ. According to the law of universal gravitation, Earth exerts a force of a larger magnitude on the more massive object (object C) than on the less massive one (object A), even if those objects are at the same location with respect to the center of Earth. However, the ratio of the magnitude of the force exerted on each object and the mass of that object is identical for all three objects:

E on A E on B E on C E2

A B C

F F F mGm m m r

" " " .

If the objects are near Earth’s surface and we substitute the values of G, mE, and the Earth’s radius rE, we find that

) *) *

24211E

22 2 6E

5.98 10 kgN m6.67 10 9.8 N kgkg 6.37 10 m

mGr

+,# $-

" , "% &' ( ,

Since this value does not depend on the mass of the object we were using as our system of interest, we can speculate that this value might be a mathematical description of the “strength” of the gravitational field near Earth. Since the gravitational force exerted on objects clearly has a direction associated with it as well, we can say that the gravitational field near Earth has a magnitude of

9.8 N kg and points directly towards the center of Earth. We call this physical quantity

characterizing the gravitational field the g! field. More precisely, let’s define the g! field at a

particular location as the net gravitational force exerted by the environment on a point-like object at that location, divided by the mass of the object. Or,

Environment on Object

Object

Fg

m"

!!

. (15.1)


Tip! The gravitational field is an idea that offers an explanation for how objects not in contact with each other can exert gravitational forces on each other. On the other hand, the g! field is one way

(there are others) of mathematically representing the gravitational field. This makes the g! field a

physical quantity (it has a numerical value with units), while the gravitational field is a way to help explain the gravitational force and is not a quantity. Electric field due to a single point-like charged object

Let’s use a similar approach to construct a physical quantity that will characterize the “strength” of the electric field. There are a few differences though. First, it’s the electric charge of an object (rather than its mass) that causes an electric field. Secondly, the electric charge of an object can be positive, negative, or zero, while the mass of an object is always positive. With these differences in mind, let’s proceed.

Imagine an object Q with positive electric charge Q , and one of three objects K, L, or M

placed at a distance r from the center of object Q (Fig.15.6a). Objects K, L, and M have positive

charges Kq q" , L 2q q" , and M 3q q" . Q will exert an electric force on K, L, or M that points in a

direction directly away from Q (Fig. 15.6b).

Figure 15.6 Q exerts electric force on one of three charged objects placed at same location

Let’s use Coulomb’s law to compare the magnitude of the electric force for each situation:

KQ on K K2 2

LQ on L L2 2

MQ on M M2 2

Qq QF k q kr r

Qq QF k q kr rQq QF k q k

r r

# $" " % &' (# $" " % &' (# $" " % &' (

.

We notice, similar to the gravitational interaction, that for objects placed one at a time at the same location, the force exerted on one of the objects is proportional to the electric charge of that object. Furthermore, if we form ratios between the magnitude of the electric force exerted on each object and the charge of that object, we find that all three ratios equal the same value:

Q on K Q on L Q on M2

K L M

F F F Qkq q q r

" " "

Since these ratios all have the same value ( 2kQ r ), it suggests that this ratio could be a

mathematical description of the strength of the electric field produced by charge Q at that location.


Since this electric field exerts an electric force on the positively charged objects K, L, or M that points in each case away from Q , the physical quantity that characterizes the electric field at a

location should be a vector that has the magnitude given above and points away from Q (if Q is

positive). We call this physical quantity the E!

field.

E!

field due to single charge Q To determine the E!

field at a specific location produced by

an object with charge Q , place another point-like positively-charged object with charge

Objectq at that location. The E!

field at that location due to Q is the ratio of electric force that

Q exerts on Objectq ( on Q qF!

) and the charge Objectq :

on

Object

Q qFE

q"

!!

(15.2)

The unit of the electric field is the newton per coulomb, or N C .

Tip! The above definition is the operational definition of the E!

field at a point, as the field created by

the object with charge Q is independent of whether we place the object with charge Objectq there or

not. As noted in the previous section, the field approach to electric force has two parts: (1) a charge

Q produces an electric field; and (2) the field exerts a force on a charge q in that field. For a single

charge Q producing a field and a single charge q placed in the field created by Q , we get the

following.

(1) To determine the magnitude of the E!

field at the location of q , use the operational

definition of the E!

field (Eq. 15.2) and divide the magnitude of the force by the charge q that

we brought in to determine that E!

:

Q on q2 2

F Q q QE k k

q r q r" " "

The result is independent of q . The magnitude of the E!

field caused by a point-like charged

object with charge Q at a distance r from that object is:

2

QE k

r" . (15.3)

cause effect


The object with electric charge Q is the cause, and the E!

field produced by it is the effect.

Note also the E!

field vector points away from the object creating the field, if Q is positive and

toward it, if it has a negative charge.

(2) We say that the E!

field at a specific location is the cause of the force exerted on a point-like object of charge q at that location. Rearranging Eq. (15.2) we can write a cause-effect relation

between the E!

field at a specific location and the force exerted on a point-like charged object q located there:

on Q qF qE"! !

(15.4)

Expressed this way, Eq. (15.4) explains why the electric field created by the same object exerts a larger electric force on an object with larger charge than it does on an object with smaller

electric charge. For example, if it exerts a force F!

on an object with charge q, it then will exert

a force 10F!

on an object with charge 10q .

We achieved our goal. We have a mathematical description for the magnitude of the E!

field a distance r from a point-like charged object and know how to determine the effect of that field on some other charge in the field. The real world is far more complex. Even in a single hydrogen atom, there are multiple charged objects (two in this case, a single proton that makes up the nucleus and a

single electron.) How do we determine the E!

field produced by multiple point-like charged objects? We investigate this question next.

E!

-field due to multiple charged objects We start by considering an example that involves two charged objects and determine the

E!

field produced by those two objects at one point in space. Example 15.1 Electric field due to multiple charged objects You have two small metal spheres each on a dielectric (non-conducting) stand. You place charge +4q on the left sphere and charge +q on the right sphere. The distance between the stands supporting the spheres is d = 1.0 m. What is the

E!

-field at the midpoint of the line connecting the spheres? Sketch and Translate The situation is sketched in Fig. 15.7a. Both charged objects (called the

environmental objects) contribute to the E!

-field at the point of interest. To solve the problem we

imagine placing a point-like positively charged system object Sq at the point of interest (the field at

that point exists even if we do not place any charged objects there). Then we determine the force that the environmental charged spheres on the left and right exert on the system object at the middle and

from that, find the E!

field at the point of interest.

cause effect


Figure 15.7(a) Determine E Field midway between +4q and +q

Simplify and Diagram Model the spheres as point-like objects so that we can use Coulomb’s law to describe the electric forces they exert on the system. A force diagram for the system is shown in Fig. 15.7b. Only electric forces are shown since our goal is to determine the electric field at that point. The electric force exerted by the left sphere on the system points in the positive x direction while the force exerted by the right sphere points in the negative x direction.

Figure 15.7(b)

Represent Mathematically Using the operational definition, the E!

-field at the point of interest equals the net electric force exerted on the system by the environment divided by the charge of the system object:

S S Snet electric on q on 4q on

S S

q q qF F FE

q q.

" "

! ! !!

.

Solve and Evaluate The E!

-field at the point of interest is a vector. We know the E!

-field there won’t have a y component; thus, we only need to determine its x component:

S Son 4 on

S

q q x q q xx

F FE

q.

"

) * ) *SS2 2 2 2 2 2

S

4 41 3 12( / 2) ( / 2) ( / 2) ( / 2) ( / 2)

k q q k qkqq kq kq kqq d d d d d d# $

" + . " + . " "% &' (

.

We see that the E!

field at the point of interest does not depend on the system object charge Sq that

we placed at that point but only on the charges on the two spheres on the left and right and the position of the point of interest.

Try yourself: What is the location of a point where the E!

field is zero? Answer: The field is zero between the source charges 0.33 m from the +q charge.


Tip! When we use the operational definition of the E!

-field, our system of interest is an imaginary point-like charged object located at the point of interest. This charged object is not real. It is part of

the reasoning process used to apply the operational definition for determining the E!

-field.

Note that in the above example, we had the expression

) *

2 2

4( / 2) ( / 2)x

k qkqEd d

" + . .

The first term on the right side is the x component of the contribution to the E!

-field of the object with charge q, while the second term on the right could be interpreted is the x component of the contribution of the object with charge 4q. In other words:

q 4q x x xE E E" . .

Apparently, when multiple charges are contributing to the E!

-field, the way to combine their contributions is to simply add them as vectors. This is known as the superposition principle.

Superposition principle The E!

-field at a point of interest is the vector sum of the

individual contributions to the E!

-field of each charged object:

1 2 3E E E E" . . .! ! ! !

" (15.5)

The Reasoning Skill below illustrates step by step how to qualitatively use the superposition principle

to estimate the direction of the E!

-field at a point of interest.

Reasoning Skill: Estimating qualitatively the E!

-Field at a point of interest (A in this example)


E!

-Field Lines

In the Reasoning Skill above the E!

-field at a single point of interest in the vicinity of three point-like charged objects was estimated. That procedure could be carried out at a hundred other

points, or a thousand, and you still wouldn’t have determined the E!

-field everywhere. There are an

infinite number of points to choose from, and at each point the E!

-field vector is usually different.

That would not be a useful way of representing the overall shape of the E!

-field. Instead, physicists

use E!

-field lines. Imagine that we have two point-like charged objects, one with positive charge q. and the

other with negative charge –q . Use the superposition principle to draw the E!

net vector at a point

equidistant from the two charges and off the axis between them (Fig. 15.8a). Then use the same

method to draw the E!

net vector for points just in front and just after that first point. Continue drawing

the E!

net vectors for adjacent points along the direction of the previous E!

net vector. You eventually

get (if done carefully) a series of E!

net vectors that seem to follow one after the other from the positive charge q. to the negative charge –q , as shown in Fig. 15.8b. Then draw a single line that passes

through each point tangent to these vectors. The line result looks as shown in Figure 15.8c. If you repeat the process for a series of points farther and close to the axis between the two charges, you can get a series of lines such as shown in Fig. 15.8d.

(a) (b)

Figure 15.8 Constructing E field line

Notice that the lines are spaced more closely (denser) near the charges causing the E!

field, places where the field has a greater magnitude and are less dense farther away where the field has a


smaller magnitude. The E!

-field lines for a single positively charged object are shown in Fig. 15.9a, for a single negatively charged object in Fig. 15.9b; and for two equal magnitude positively charged objects in Fig. 15.9c.

Figure 15.9 E field lines

Summary— E!

-field lines:

! The E!

-field lines are drawn so that the E!

-field vectors at points on those lines are tangent to those lines.

! The direction of the E!

-field line at a point is the same as the direction of the E!

-field vector at that point.

! E!

-field lines begin on positively charged objects and end on negatively charged objects.

! The number of E!

-field lines beginning/ending on a charged object is proportional to the magnitude of that object’s charge.

! The density of the E!

-field lines in a region is proportional to the E!

-field vectors in that region.

Tip! Despite their suggestive appearance, E!

-field lines DO NOT necessarily represent the trajectories that charged objects take when passing through the field. They do give information about the direction of the force the electric field exerts on a charged object, but that’s connected with the charged object’s acceleration, not its velocity (Newton’s 2nd law.) Review Question 15.2

How do you calculate the E!

-field at a point located near two point-like charged objects?

15.3 Skills for analyzing processes using E!

-Field Determining the E

!-field at a point of interest allows us to easily determine the electric force

the field will exert on a charged object located at that point:

O on OEF q E"!! !

.


Then, we can use Newton’s 2nd law to determine its acceleration, which then lets us determine the object’s motion. In this section we develop the skills needed to do this.

Determining the E!

-field produced by multiple point-like charged objects

The following Reasoning Skill describes the procedure for determining quantitatively the E!

-field at a point.

Reasoning Skill: Calculate the E!

Field Follow the steps shown below to determine the magnitude

and direction of the E!

-field at position I.

Example 15.2 E

!-field due to multiple charges Take a metal ball, attach it to a plastic stand and

place it on a table a distance d from a second ball/stand. Use a Wimshurst generator to give one of the

balls a positive charge +q and the other a negative charge –q. Determine an expression for the E!

-field at a distance d above the center of the line connecting the objects.


Sketch and Translate The sketch in Fig. 15.10a shows the charged objects +q and –q that are

producing the E!

-field, and shows the location of interest (the point where we want to determine the

E!

-field.) Since two charges are contributing to the field, we need to use the superposition principle.

To do this we place an imaginary positively charged object Sq at that location of interest. It is the

system object.

Figure 5.10(a) Determine E field

Simplify and Diagram Next, construct an E!

-field diagram for this point of interest. Draw an E!

-field

arrow for each charged object that contributes to the E!

-field at that point (Fig. 15.10b.) Make the relative size of the arrows consistent with each other.

Figure 15.10(b)

Represent Mathematically Use Eq. (15.4) to determine the magnitude of the E

!-field contribution

from both +q and –q:

2q qqE E kr. +" "

To determine the distance r from +q and –q to the location of interest, use the Pythagorean theorem: 2

2 2 2

2 2

5( )2 4

52

455

2

q q

d dr d

dr

q kqE E kdd

. +

" . "

/ "

" " "# $% &' (


Next, the angle 0 that the E!

-field vector makes with the positive x-axis is:

tan y

x

EE

0 "

Now determine the x and y components of the E!

-field. Look carefully at Fig. 15.10a and carefully pay attention to how cos0 is determined. First the x component:

2 2

4 2 8cos cos 2 cos 25 5 2 5 5x q x q x q q q

kq d kqE E E E E Ed d d

0 0 0. + . + .

# $# $" . " . " " "% &% &' (' (

.

And now the y component:

y y sin sin 0y q q q qE E E E E0 0. + . +" . " . + "

Notice that the y components of the two E!

-field contributions cancel. This happened because of what is known as the symmetry of the situation. The magnitudes of +q and –q are the same, and the distances from them to the point of interest are the same. Thus, their y components have equal magnitudes, but one pointing in the +y direction and the other in the -y direction; so that they add to zero.

We can now determine the direction of the E!

-field. Since its y component is zero, it must point in the positive x direction. Let’s check:

) *2

1

0tan 085 5

tan 0 0

y

x

EkqE

d

0

0 +

" " "

/ " " 1

Solve and Evaluate The magnitude of the E!

-field is

2

2 2 22 2

8 805 5 5 5x y

kq kqE E Ed d

# $" . " . "% &' (

and it points at an angle of ) *1tan 0 00 +" " 1 counterclockwise from the positive x direction,

meaning that it points exactly in the positive x direction. The units are appropriate (using the above equation for E):

2

2

2

N m CC Nm C

# $-% &' ( "

Try It Yourself: Determine the E!

-field at point I in Fig. 15.11. (a) Calculate the magnitudes of the

E!

-field contributions from each of the charged objects at point I. (b) Calculate the x and y

components of those E!

-field contributions. (c) Determine the net x and y components of the E!

-field

at point I. (d) Finally, determine the magnitude and the direction of the E!

-field at point I.


Figure 15.11

Answer: (a) 225 N, 144 N; (b) (225 N, 0) and ( –115 N, +87 N); (c) +110 N, +87 N; (d) 0140 N, 38 above x. axis.

Producing a uniform E!

field We see that even in situations with a very small number of charged objects contributing to the

E!

-field, the shape of the field is very complicated. Its magnitude and direction are different at every single point. Charged objects that are free to move in this region will move in very complicated ways since the electric force exerted on them will be continuously changing as they move from point to

point. This suggests a question: Is it possible to arrange charged objects in such a way so that the E!

-field they produce in a region will be the same at each point (have the same magnitude and point in

the same direction), a so-called uniform E!

-field? If possible, charged objects moving in this region would have constant acceleration, a kind of motion we are able to mathematically describe using kinematics. This would seem like an impossible search since there are an infinite number of ways to position charged objects. However, there is a very common everyday example of a vector field that is approximately uniform. Near the Earth’s surface, g! -field has approximately constant magnitude

pointing vertically downward. Since that configuration of mass seems to produce a uniform g! -field,

perhaps an analogous configuration of charge will produce a uniform E!

-field. The next exercise investigates this possibility.

Conceptual Exercise 15.3 Uniform E!

-field Draw E!

-field lines for a large uniformly charged plate of glass. Sketch and Translate The plate of glass is shown in Fig. 15.12a. “Uniformly charged” means that the amount of electric charge located on each unit area of the surface is constant throughout the plate. This can occur if you rub the plate of glass evenly with silk (which removes electrons from the plate leaving it positively charged.) Figure 15.12(a) E field due to

large charged plate


Simplify and Diagram Model the plate of glass as being infinitely large in both perpendicular

directions in the plane of its surface. To understand the shape of the E!

-field lines produced by the

plate, choose a point of interest to the right of the plate in the region near its center. Think of the E!

-field contributions from each of the excess positive charges on the plate (Fig. 15.12b) at this point of

interest. The plate charges farther from this point provide smaller contributions to the E!

-field. Also,

the y component of the E!

-field contribution from one of the plate’s charges (say at position 1 in Fig. 15.12b) is canceled by the y component of the contribution from a plate charge at some other point

(position 2). This occurs for every pair of plate charges. Therefore, the y component of the E!

-field is

zero at every point in the region near the plate. Consequently, the E!

-field is perpendicular to the

plate at every point in that region. (Fig. 15.12c) Since E!

-field lines are parallel to the field at every

point, the E!

-field lines must point away from the plate and be perpendicular to it. (Fig. 15.12d)

Figure 5.12(b)(c)(d)

Since the E!

-field lines are parallel to each other, are perpendicular to the plate, and there are

no charged objects outside the plate to produce additional E!

-field lines, the field lines must have a

uniform density—the same separation between the field lines everywhere. Since the density of E!

-

field lines represents the magnitude of the E!

-field, this means the E!

-field has the same magnitude

everywhere in this region. The uniform charge on the plate produces a uniform E!

-field. Try It Yourself: Imagine that an infinitely large uniformly positively charged plate oriented in the

plane perpendicular to the page produces an E!

-field of magnitude E . You add another plate, negatively charged, at a distance d from the first one and parallel to it. What is the magnitude of the

E!

- field to the left of the plates, between them, and to the right? Answer: 0, 2 , 0E .

Incorporating the electric force into Newton’s second law

In Chapters 2 and 3 we learned to use Newton’s second law to relate the forces exerted on a system to the resulting acceleration of the system. Now that we know how to represent the electric


interaction as a force ( E.f. on O OF q E"! !

), we can use Newton’s second law to determine the motion of

charged objects that have electric forces exerted on them. A general problem solving strategy for doing this is described and illustrated in Example 15.4.

Example 15.4 A charged object in a known E!

-field A spring made of a dielectric (electrically insulating) material with spring constant 50 N/m hangs from a large uniformly charged horizontal

plate. The uniform E!

-field produced by the plate has magnitude 52.0 10 N C, and points down. A

0.20-kg ball with charge 54.0 10 C+. , hangs at the end of the spring. Determine the distance the

spring is stretched from its equilibrium length. Assume that the gravitational constant is 10 N/kg .

Sketch and Translate • Draw a labeled sketch of the situation. Include the symbols for the known and unknown quantities that youplan to use. • Choose a system of interest.

The situation is sketched at the right. The charged ball isthe system of interest.

Simplify and Diagram • Determine the E

!-field produced by the environment.

Is it produced by pointlike charges (making it non-uniform) or by large charged plates (making it uniform)? • Construct a force diagram for the system. • State any assumptions you have made.

The field is uniform. Assume that the spring has no mass. A force diagram for the system is shown at the right with arrows drawn to scale.

Represent Mathematically ! Use the force diagram to help apply Newton’s second law in component form. Use component addition to determine the net force along each coordinate axis. Determine the acceleration of the system if needed. ! If necessary, use kinematics equations to describe the motion of the system.

The ball is not accelerating so the forces exerted on theball balance. None of the forces have x components:

S on B E on B Plate on B

( ) ( ) 0

0y y y

k x mg qE

mg qEx

k

F F F/ 2 . + . + "

./ 2 "

. . "

Solve and Evaluate ! Combine the above equations and complete the solution. ! Check the direction, magnitude, and units, and decide whether the result makes sense in limiting cases.

5 5(0.20 kg)(10 N/kg) (4.0 10 C)(2.0 10 N/C)

50 N/mx

+. , ,2 "

0.20 mx/ 2 " The units for the stretch distance are meters, as they should be. Let us check it for the limiting cases. As boththe gravitational force and the electric force point in the same direction, eliminating either of them should reducethe distance that the spring stretches. Note in the


equation above that x2 decreases if 0m " or if 0q " . Also, a stiff spring (larger k ) should stretch less if we have the same mass ball and the same electric charge – also consistent with our result.

Try It Yourself: Suppose the plate was negatively charged producing the same magnitude electric field but now pointing up. Now, what would the spring stretch or compression be? Answer: –0.12 m (the spring is compressed 0.12 m).

Example 15.5 An object moving in an E!

-field Inside an inkjet printer a tiny ball of black ink of

mass 1.9100 g3 with charge 96.0 10 C++ , moves horizontally at 40 m/s . The ink ball enters an

upward pointing uniform E!

-field of magnitude 41.0 10 N C+, produced by a negatively charged

plate above and a positively charged plate below. The plates are used to deflect the ink ball so that it lands at a particular spot on a piece of paper. Determine the deflection of the ink ball after it travels

0.010 m in the E!

-field. Sketch and Translate After sketching the situation (Fig. 15.13a), we choose the charged ink ball as the system. The electric field due to the plates exerts a downward force on the negatively charged ball. In addition, the Earth’s gravitational field also exerts a downward force on the ball. No other objects or fields interact with the ball. We break the problem into two parts: first determine the acceleration of the ball due to the forces being exerted on it and then use kinematics to determine its vertical displacement as it passes through the region with the electric field.

Figure 15.13(a) Charged ink ball deflected by electric field

Simplify and Diagram A force diagram for the ball is shown in Fig. 15.13b. We model the ball as a

point-like object and the plates as producing a uniform E!

-field.

Figure 15.13(b)


Represent Mathematically The forces exerted on the ink ball do not have x components, so we use only the y component form of Newton’s second law (we choose the positive y direction pointing down):

) * ) *P on B E on B 1 1 1

y y y y y ya F F F qE mgm m m

" " . " .4

Here, yE is the y component of the E!

-field caused by the electric charge on the plates and yg is the

y component of the g! -field caused by the Earth. Both are positive with our choice of coordinate

system.

yy y

qEa g

m/ " .

The ink ball does not accelerate in the x direction, so the x component of its velocity will remain

40 m/s during the time it takes to traverse the region with nonzero E!

-field. This time interval is

xt x v2 " 2 . During this time interval the ball’s displacement in the y direction is:

20 0

12y yy y v t a t+ " 2 . 2

Solve and Evaluate Substituting for ya , t2 and 0 0yv " gives:

2

0102

yy

x

qE xy y gm v

# $# $ 2+ " . . % &% &

' (' (

) *) * 29 44

9

6.0 10 C 1.0 10 N C1 0.010 m9.8 N kg 1.9 10 m 0.19 mm2 100 10 kg 40m s

+ ++

+

# $+ , , # $% &" . " , "% &% &, ' (' (

.

The units in the answer are the units of distance and the value seems reasonable given the size of a letter on a printed page. Notice that we can move the place that the ball hits the screen by changing

the magnitude and/or direction of the E!

-field. This is the principle behind how an inkjet printer works. Notice also that the gravitational force is negligible compared to the electric force and has essentially no effect on the motion of the ball.

Try It Yourself: What should the mass of the ink ball be so that if you reverse the direction of the E!

-field, the ball will have zero acceleration?

Answer: 66.0 10 kg+, .

Review question 15.3

What is the difference between the E!

-field produced by a point-like charged object and the E!

-field produced by a large uniformly charged plate?


15.4 The V-field

We operationally defined the E!

-field at a specific location as the net electric force exerted by the environment on a positively charged point-like object located there, divided by that object’s charge:

Environment on Object

Object

FE

q"

!!

.

The magnitude of the E!

-field produced a distance r from a charge Q is

2

kQEr

" .

The E!

-field vectors of a single positively charged object at any location point away from it and

toward a negatively charged object. The E!

-field produced at any point by multiple environmental

objects is the vector sum of the E!

-fields produced by each environmental charged object. If a system

object with charge Oq happens to be in that E!

-field, the field exerts an electric force on the object:

E.f. on O OF q E"! !

.

This allows us to incorporate electric forces into Newton’s second law, which then allows us to predict the motion of charged objects affected by the electric field. However, while studying mechanical phenomena, we found that using a work-energy approach was often an easier way to analyze certain processes. Now our goal is to incorporate the electric interaction into that work-energy approach. In order to do this we need to describe the electric

field not as the force-related E!

-field, but instead as an energy-related field. To construct this new way of representing the electric field, we think of the electric

interaction in terms of the electric potential energy of a system. Let’s imitate how we constructed

the quantity of E!

-field. Consider the electric potential energy of interactions of a point-like electrically charged object with charge Q and the charged objects K, L, or M with charge

K L M, , or q q q (all with different electric charge) separated by a distance r from Q (Fig.

15.14).

Figure 15.14 Energy of system with Q and q = qR, qL or qM

Recall from Chapter 14 that the electric potential energy of two point-like charged objects separated by a distance r is:


1 2

1 2q q

kq qUr

" .

Thus, the electric potential energy of charges K L M and , , or Q q q q is:

K

Kq Q K

kq Q kQU qr r

# $" " % &' (

L

LLq Q

kq Q kQU qr r

# $" " % &' (

M

MMq Q

kq Q kQU qr r

# $" " % &' (

If we consider ratios between the electric potential energy of the Q - K L M, , or q q q charge pair

and the charges K L M, , or q q q (that is /qQU q ), we find that all three ratios equal the same

quantity:

K L M

K L M

q Q q Q q QU U U kQq q q r

" " " .

Since these ratios all have the same value ( kQ r ), it suggests that this ratio could be a mathematical

representation of the electric energy field caused by charge Q at a distance r from Q . We call this

physical quantity the V field or electric potential due to Q .

V-field (or electric potential) The V-field is a physical quantity that characterizes the energy (as opposed to the force) of the electric field at a specific position. It equals the total electric

potential energy QqU of the system consisting of an imaginary positively charged point-like

object q located at that position and the objects creating the field divided by the charge of

the object with charge q :

all QqUV

q" (15.6)

The unit of the V-field is called the volt (V) where 1 V = 1 J/C (joule/coulomb).

The V-field is another way of representing the electric disturbance that charged objects produce in the regions surrounding themselves; but in this case, it’s connected with ideas of work-energy instead of forces. Using this operational definition of the V-field, we can say that the

electric potential energy qU of a system of a point-like charge Oq and all charged objects

creating the V-field at Oq ’s location is:

OqU q V" (15.7)


Tip! Unlike the E!

field, which is a vector, the physical quantity V-field (electric potential) is a scalar. Thus, it can have a positive or a negative value depending on the signs of the electric charges of the objects that create the V-field at a particular location. Electric potential due to single negatively-charged object

Let’s use what we have been learning to analyze the V-field of a single negatively-charged point-like object at different distances from the object.

Quantitative Exercise 15.6 Electric potential produced by a negative point-like charge You have a point-like charged object with negative charge Q+ . Determine the value of the V-field at points that

are distances a , 2a , and3a from this object (Fig. 15.15a). Then, construct a V-field-versus-distance graph for the V-field produced by that object.

Figure 15.15(a) Electric potential V due to single negative charge –Q

Represent Mathematically To determine the V-field at a specific location, we first imagine that there is a point-like positively charged object q located at that location. We then divide the electric potential energy of the interaction of that object with the other charged objects objects by q. There is one other object present and it has charge Q+ .

At a distance a from –Q: – at a– at a

(– ) (– )QqQq

UQ q QU k V ka q a

" / " " .

At a distance 2a from –Q: – at 2a– at 2a

(– ) (– )2 2

QqQq

UQ q QU k V ka q a

" / " " .

At a distance 3a from –Q: – at 3a– at 3a

(– ) (– )3 3

QqQq

UQ q QU k V ka q a

" / " " .

It seems that the V-field as a function of the distance r from object –Q is: (– ) QV k

r" .

The function is plotted in Fig. 15.15b. We use a coordinate system with its origin at the location of the object with charge Q+ . Notice that the V-field becomes more and more negative as get closer

Q+ and slowly becomes less negative with increasing distance from Q+ . This means that if an

object with positive charge q is located close to object –Q and then moved away from it, the electric


potential energy between the two charges qU qV" would increase (become less negative). This

makes sense, as we need to pull a positively charged object away from a negatively charged one. In other words, positive work must be done to the system (Fig. 15.15c) to separate the oppositely charged objects.

Figure 15.15(b)(c)

Try It Yourself: Repeat this example only for a positive source charge +Q.

Answer: The potential will be + QV k

r" and the potential varies with distance from the source

charge as shown in Fig. 15.16.

Figure 15.16 Electric potential-versus-r for +Q

In general, a point-like charged object with charge Q produces a V -field at the distance r from itself

given by the expression:

QV kr

" . (15.8)

When using the above, we must include the sign of the charge causing the -fieldV . Tip! It is important to remember that potential energy depends on the choice if the location where it is decided to be zero. Unless we specify otherwise, we will always assume that the electric potential energy of the system of two objects is zero when the distance between them is infinity – they are so far away from each other that they do not interact.


Comparing the E!

field and the potential field V

Compare the equation for the magnitude of the E!

-field due to a point-like charged object with charge Q at a distance r from the object

2

QE k

r"

and the -fieldV due to that same charged object at that same distance r from the object:

QV kr

" .

They are similar in that they both depend on the charge of the object and on the distance from that object to the location of interest. However, there are differences. First, since E is the magnitude of the

vector E!

-field, it is always positive regardless of the sign of the object’s charge (the absolute value in the equation guarantees this). On the other hand, V can be either positive or negative depending on

the sign of the charge. Secondly, E is proportional to 21 r whereas V is proportional to 1 r . This

means that E decreases with distance from the object more quickly than V does. The V-field and the superposition principle

Now that we can determine mathematically the V-field produced by a single point-like charged object (Eq. 15.8), we can use the superposition principle to determine the V-field at a specific

location produced by several charged objects. Using the same idea as for the E!

-field we have:

1 2 3V V V V" . . ." (15.9)

Because the V-field is a scalar field rather than a vector field (like E!

-field is), it is much easier to apply the superposition principle. Just as in mechanics, the energy approach (representing the electric field with the V-field) is often easier than the force approach (representing the electric field with the

E!

-field.) Example 15.7 V-field in body due to dipole charge on heart At one instant during a human heartbeat, the heart has the dipole charge distribution shown in Fig. 15.17. A dipole charge distribution is comprised of one point-like positively charged region, and one negatively charged point-like region of the same magnitude. (a) Determine the V-field at points A and B inside the body beside the heart. (b) Would a positive sodium ion Na+ tend to move from A to B or from B to A? Would a negative chlorine ion Cl– tend to move from A to B or from B to A? Explain.


Figure 15.17 Potential V at A and B due to heart dipole charge

Sketch and Translate The situation with the known information is sketched in Fig. 15.17. We need to think about whether ions (positive and negative) accelerate toward lower or higher V-field regions (toward lower or higher potential). Simplify and Diagram We assume that the charged regions of the heart are point-like, and that the body tissue does not contribute to the V-field. Represent Mathematically Use the information in Fig. 15.17 and Eq. (15.8) (V-field produced by a single point-like charged object) and Eq. (15.9) (superposition principle) to determine the electric potential at A and B due to the heart dipole charges

A – 1 1 + = + (– )

0.15 m 0.17 m 0.15 m 0.17 mq qV k k kq # $" % &

' (

B – 1 1 + = + (– )

0.12 m 0.14 m 0.12 m 0.14 mq qV k k kq # $" % &

' (

Solve and Evaluate Substituting the known information into the above, we get:

9 2 2 –7A

1 1 (9.0 x 10 N m /C )(4.0 x 10 C) – = 0.028 V0.15 m 0.17 m

V # $" , % &' (

.

9 2 2 –7B

1 1 (9.0 x 10 N m /C )(4.0 x 10 C) – =0.043 V0.12 m 0.14 m

V # $" , % &' (

.

For part (b), consider an analogy to the gravitational interaction. Imagine an object-Earth system. First you are holding the object and then you let it go so the only force exerted on it is the gravitational force. Independently of where we choose the zero level of the potential energy of their interaction (at the surface of the Earth or at infinity), the object will start accelerating towards Earth, making the gravitational potential energy of the system less. This idea that systems evolve in such a way that tends to reduce their gravitational potential energy applies to the electric interaction as well. Therefore, a positively charged particle (like a sodium ion) will accelerate towards regions with lower V -field—from B to A. The positive ion’s electric potential energy is lower at A than at B. A negatively-charge particle (like a chlorine ion) will accelerate towards regions with higher V -field—from A to B. The negative ion’s electric potential energy is lower at B than at A.


Try It Yourself: Determine qualitatively the direction of the net electric force that the dipole charges exert (a) on a positive sodium ion half way between A and B and (b) on a negative chlorine ion that is half way between A and B. Is this consistent with the idea that a charged object tends to move toward regions of where the electric potential energy of the system is lower ? Explain. Answer: (a) The net force that the heart’s dipole charges exert on a positive sodium ion is toward A (the positive ion of the heart dipole is closer to the sodium ion and exerts a stronger repulsive force than the attractive force exerted by the negative ion on the sodium ion). Note also that the heart dipole- sodium ion system has less electric potential energy at A. (b) The net force that the dipole charges exert on a chlorine ion is toward B (the heart’s positive ion is closer to the chlorine ion and exerts a stronger attractive force than the repulsive force exerted by the heart’s negative dipole charge). Note also that the heart dipole-chlorine ion system has lower electric potential energy at B. Electrocardiogram Our heart contracts in a rhythmical way pumping blood through the body. How does it know when to contract? The rhythm of the heart is controlled by the nerve cells that produce electrical impulses called action potentials that trigger the electrical activity of the heart. The electrical action potentials come from cell membranes where chemical processes cause different concentrations of positive and negative ions on the two sides of the membranes. This is an example of charge separation (polarization) and produces a V-field that varies from one side of the membrane to the other (known as a potential difference.) When some event triggers a depolarization of the membrane, the potential difference across it changes. Such changes occur in different muscle cells in the heart and collectively cause different parts of the heart to contract at different times in the heart beat cycle.

The charge separation in the heart is what an electrocardiogram measures. In other words, it measures the potential difference (the difference in the value of the V -field) produced by the dipole charges of the heart. The dipole charges depend on which heart muscle cells are compressing at a particular time. For example, during a person’s left ventricle contraction (the main pump for the heart), a large number of muscle cells are contracting and the dipole charge distribution is greater than during left ventricular contraction when blood is pumped from the heart to the lungs. The more muscle cells contracting, the greater the heart’s electric dipole and the greater the difference in the V -field between different parts of the body away from the heart. If the heart has to work harder than usual because of clogged arteries or a congested peripheral circulatory system, then the heart muscles contract more when pumping the blood and the dipole charge is larger than in a healthy heart. The larger dipole charge produces a larger potential difference, which is easily measured by an electrocardiogram. We will return to the details of the electrocardiogram in the last section of this chapter.

The V-field or electric potential will be a very useful physical quantity in understanding electric circuits in Chapter 16. It is also useful in analyzing processes where the electric interaction is used to accelerate charged objects such as in television sets, x-ray machines, and particle accelerators.


Review question 15.4 How do you determine the V-field at a chosen location of interest?

15.5 Equipotential lines—representing the V-field We’ve used E

!-field lines to help visualize the E

!-field. In this section we will learn a new

graphical representation for the V-field that helps us visualize it. Examine the V-field at different points in the region surrounding the positive point-like charged object +Q shown in Fig. 15.18. The value of the V -field at points A, B, and C is the same as at point 1, and its value at D, E and F is the same as at point 2. If we connect these points of equal V-field value, we get a pair of what are called V -field lines, also called equipotential lines. These are lines on which the V -field has the same value at every point. In this case, the lines are circles because the V -field produced by a point-like charged object has the same value at all points equidistant from the charged object.

Figure 15.18 Equipotential lines

Since the value of the V -field is constant along an equipotential line, it means that if a second positively charged point-like object +q is placed anywhere on that line, then the electric potential energy of the two-charge system will be the same regardless of where on the line +q is located. Furthermore, it means that no work needs to be done on +q by an external force to move it

along the line. This makes sense because the E!

-field points away from Q. . In order to move q.

from one point on the line to another, an external force need only be exerted on q. pointing towards

Q. in order to balance the electric force that Q. exerts on q. Since this external force will be

exerted in a direction perpendicular to any direction along the equipotential line, the work done on the

two-charge system by the external force will be zero (since 0cos cos90 0W Fd Fd0" " " ).

This brings up an important relationship between E!

-field lines and equipotential lines. In the previous two paragraphs we learned that the V-field of a point-like charged object could be


represented by spherical equipotential lines centered on the object. We also remembered that the E!

-

field of that same object can be represented with E!

-field lines that point radially away from the

object. This means that the equipotential lines and the E!

-field lines are everywhere perpendicular. If

an electric field is created by a uniformly charged infinitely large plate, its E!

-field lines are parallel to each other and perpendicular to the plate (Fig. 15.19). If the equipotential lines of this field are

perpendicular to the E!

-field lines, they should be parallel to the plate, as shown in the figure.

Figure 15.19 E field lines and constant V surfaces for large charged plate

In Fig. 15.20a we draw again the E!

-field lines and equipotential lines for the field created by a positively charged point-like object Q. . A graph of the V-field -versus-distance from Q. (V-

versus-r) is shown in Fig. 15.20b. In (a) the value of the V-field on adjacent equipotential lines differs by equal amounts. Notice that the lines that represent the values of potential that differ by the same amount are closer together when nearer Q. and further apart when farther away from Q. . The

closer to Q. region the more rapidly the V -field changes with changing distance r. This is also

where the density of the E!

-field lines is greatest. Thus in regions where the E!

-field is stronger, the

V-field changes more rapidly with position. This also means that in the areas where E!

field is zero, the value of the V field is the same at all locations, or the whole region is one big equipotential region!

Figure 15.20 Variation of E and V for positive point charged object


Remember that an object interacting with Earth only accelerates in a direction that will decrease the gravitational potential energy of the object-Earth system. As an example, an object that is thrown upwards has a downward acceleration, which is the direction of decreasing gravitational potential energy of the object-Earth system. Charged objects that are interacting electrically follow a similar pattern, accelerating in a direction that decreases the electric potential energy of the system. The situation with electrically charged objects is somewhat more complicated since the electric charge of an object can be either positive or negative while the mass of an object can only be positive. Let’s investigate this.

Looking at Figs. 15.19 and 15.20 we can see that the E!

-field points perpendicular to the equipotential lines and toward regions of lower V-field value. This means that a positively charged object accelerates from regions of higher V -field value toward regions of lower V -field value. A negatively charged particle tends to do the opposite, accelerating from regions of lower V -field value toward regions of higher V -field value. In either case, the change in electric potential energy of the system is

) * f i f i f iq q

q

U U qV qV q V VU q V+ " + " +

/ 2 " 2.

This change in electric potential energy can be positive or negative depending on the sign of the

charge q that is moving, and the V -field values ( iV and fV ) at its initial and final locations.

If this discussion of V -field lines seemed abstract and difficult to relate to, here’s an analogy that might help. You probably recall contour maps with lines that indicate constant elevation above sea level (Fig. 15.21). Each line represents a set of points that are at a specific elevation. If you were hiking in the mountains or riding a bicycle in the Tour de France, you would move up and down between different elevations. As a result, the gravitational potential energy of the you-Earth system changes. However, at all points along a contour line, that gravitational potential energy would be the

same value, gU mgy" where y is the elevation of that contour. If we divide this by the mass of the

object, we get a quantity that is known as the 5 -field, gUm

5 " (that’s the lowercase Greek letter

‘phi’). This field is also known as the gravitational potential and is an alternative way of representing the gravitational field (the other way being the force-like g! -field.) The object that is producing this

5 -field is Earth. Just as the V -field can be represented by V -field lines, the 5 -field can be

represented by 5 -field lines, and these lines are what you see on a contour map.

The V-field is then analogous to the5 -field, which makes it a sort of ‘electric elevation.’ It’s

not a real elevation, but you can think of it like one. Thus the contour lines are similar to the equipotential lines in the electric field. You also know that in the regions where the contour lines are closer together, the elevation changes faster with the position; the same it true for equipotential lines.


Figure 15.21 Constant elevation contour lines

We learned that positively charged objects accelerate from regions of higher V -field value toward regions of lower V -field value. Referring to the analogy, this would be the ‘downhill’ direction. For negatively charged objects the analogy is a bit strange. Imagine what would happen if an object existed with negative mass. The gravitational field would exert a force on this object in the opposite direction. It would accelerate upward away from Earth! Physicists do not have any evidence for the existence of negative mass objects, but there most certainly exist objects with negative electric charge. This means that negatively charged objects will accelerate in the ‘uphill’ direction in our analogy, from regions of low V -field value to high V-field value. Review question 15.5 You want to move a small positively charged object in a circular path around a charged aluminum foil ball. What work is done when you move 1/4 of the circumference compared to moving 1/2 of the circumference (you are moving the charged object very slowly, so disregard its change in kinetic energy)?

15.6 Skills analyzing processes using V-Field approach The skills for using work-energy ideas to analyze processes involving the electric interaction

are similar to the skills learned in Chapter 6 for analyzing mechanical processes. These skills are described and illustrated in the following problem.

Example 15.8 X-ray machine In an X-ray machine there is a wire (called a filament) which, when hot, ejects electrons (we will learn more about this process in Chapter 26). Imagine one of those electrons that is now outside the wire, starts at rest and accelerates through a region where the V-field increases by 40,000 V (also called a +40,000 V potential difference.) The electron stops abruptly when it hits a piece of tungsten at the other side of the region, producing X-rays (more about that in Chapter 26.) How fast is the electron moving just before it reaches the tungsten?


Sketch and Translate • Sketch the initial and final states of the process described in the problem statement. Include symbols for the relevant known and unknown quantities. • Choose a system of interest. This will usually include the electric field. •Choose a zero-level for the V-field if necessary.

The situation is sketched below.

The system is the electron and the electric field of the tube. We choose the zero level of the V-field to be at the initial position of the electron, just outside the filament.

Simplify and Diagram • Identify your assumptions. • Construct a work-energy bar chart representing the process.

Assume the gravitational force that the Earth exerts on the electron is not significant. The bar chart is shown below. In the initial state, the system has neither electric potential energy (the V-field is zero there) nor kinetic energy (electron is not moving). In the final state, the system has positive kinetic energy and negative electric potential energy. The electric potential energy is determined using Uq = qV, with 0q 6 and 40.000 VfV " . in the final situation.

Represent Mathematically • Use the bar chart to help apply the generalized work-energy equation for the process.

Each non-zero bar in the bar chart turns into a term in the generalized work-energy equation.

210

2 e f e fm v q V" .

Solve and Evaluate • Solve the above equation(s) for the unknown quantity. • Check the magnitude and units, and decide if the result makes sense in limiting cases.

2f

e f

e

q Vv

m" +

19 48

312( 1.6 10 C)(4.0 10 )

1.2 10 m/s9.11 10 kg

V+

+

+ + , ," " ,

,.

Let’s check the units:

2C J/C J N m kg m m m

kg kg kg s sfvkg, , , ,

" " " " ",

.

We get the correct units for speed. Limiting case check: If the electric charge of the electron were zero, it would not participate in the electric interaction at all and its final speed should be zero; it is. If the V-field in the final state were the same as in the initial state, the final speed of the electron should be zero; and it is.


Try It Yourself: A 0.10-kg cart with a charge of +6.0 x 10-5 C rests on a 37o incline. Through what potential difference must the cart move along the incline so that it travels 2.0 m up the incline before it stops? The cart starts at rest. Assume that g = 10 N/kg. Answer: The potential difference needed is +20,000 V. Review Question 15.6

If you place a very light positively charged object in a uniform E!

-field where the V-field value decreases from left to right, in which direction will it accelerate? If the object were negatively charged instead, in which direction will it accelerate?

15.7 Relating the E!

Field and V -field We already know that E

!-field lines point in the direction of decreasing V -field and the E

!-

field lines are perpendicular to the V -field lines. It is possible to construct a quantitative relationship

between the E!

-field and V-field as well? We know that the V -field varies most rapidly with position

where the E!

-field is strongest. To make this relationship quantitative, consider the uniform E!

-field

produced by an electrically charged infinitely large metal plate. The E!

-field lines are perpendicular to the surface and equally spaced. If we place a small object with charge +q near the plate, the plate’s electric field exerts a force on the object pointing away from the plate (Fig.15.22a). We now attach a very thin wooden stick to the charged object. The stick exerts a force on the charged object pointing towards the plate (Fig. 15.22b) that balances the force exerted by the electric field on the object

( S on O P on O 0F F. "! !

). (Fig.15.22c)

Figure 15.22 Experiment to relate E and V change


Now, let’s do a work-energy analysis for a process when the charged object (still attached to the stick) is allowed to move slowly a small distance !x away from the plate (Fig. 15.22d.) We choose the charged object and the electric field as the system of interest, but not the stick, which is part of the environment and does work on the system. The only energy change is the system’s electric potential energy as the positively charged object moves farther away from the positively charged plate (see the bar chart at 15.22e). Applying the generalized work-energy equation, we get:

i fq q

q

U W UW U. "

/ " 2

The stick exerts a force on the charged object opposite its displacement. This means the work done by the stick on the system is negative:

) *S on O S on Ocos 180W F x F x" 2 1 " + 2

Applying the x-component form of Newton’s second law to the force diagram in Fig. 15.22c and

noting that P on O S on O 0F F. "! !

, we get:

P on O S on O 0x xF F. "

S on O(– ) 0xqE F/ . "

Thus, the magnitudes of the forces are equal ( S on O xF qE" ). Substitute this into the expression for

the work done by the stick on the system: S on O– xW F x qE x" 2 " + 2 . We found earlier that

qW U" 2 . We also know that qU q V2 " 2 . Thus, we get finally:

q xW U q V qE x" 2 " 2 " + 2

This leads to a relation between the potential difference between two points and the component of the

E!

-field along the line connecting those points:

xV E x2 " + 2 (15.10)

Equivalently, the component of the E!

-field along the line connecting two points is proportional to how much the V -field changes between those two points:

xVEx

2" +

2 (15.11)

The magnitude of the E!

-field component in a particular direction indicates how fast the V-field changes in that direction. If the V -field does not change in that direction it means that the

magnitude of E!

-field component in that direction is zero. Notice that the vector component of the

E!

-field points in the direction of decreasing V -field, hence the minus sign in Eq. (15.11). Similar equations apply for other directions if the situation is in two or three dimensions.

Tip! Equation (15.11) shows that the E!

-field unit N/C is equivalent to V/m.


Although we derived Eqs. (15.10) and (15.11) using the example of a uniform E!

-field, they

represent a general result that relates the component of the E!

-field in the chosen direction and the

rate of change of the V -field in that direction. However, if the E!

-field is not uniform (for example, in the region surrounding a point-like charged object), we must choose two points that are very close to each other when applying Eqs. (15.10) and (15.11).

Conceptual Exercise 15.9 Can you think of charge distributions and locations relative to those

charges where: (a) the E!

field is zero but the V -field is not zero; and (b) the V -field at a particular

location is zero but the E!

field is not. Sketch and Translate See the examples in Fig. 15.23a and b. Consider points that are in the exact centers between the charges.

Figure 15.23(a)(b) E and V fields

Simplify and Diagram (a) Note that at the point between the charges in Fig. 15.23a, the electric field points toward the right (see Fig. 15.23c) but the potential is zero:

( ) (– ) 0/ 2 / 2

k Q k QVd d.

" . " .

(b) Note that at the point between the charges in Fig. 15.23b, the electric field is zero (see Fig. 15.23d) but the potential is:

( ) ( ) 4 ( )/ 2 / 2

k Q k Q k QVd d d. . .

" . " .

Figure 15.23(c)(d)

Try It Yourself: Suppose four equal positively charged objects are at the corners of a square. Is there anyplace in the plane of the square where either of the conditions (a) or (b) described in the last conceptual exercise is met? Explain.

Answer: At the center of the square, the electric E!

-field is zero but the electric -fieldV is positive.


Testing the relationship between E!

-field and V2

We derived Eq. (15.10) using our knowledge of work-energy, the E!

-field, and the -fieldV . Is it possible to test this relationship experimentally? To do this, we need instruments that can record

and measure the value of the E!

-field and the -fieldV at specific points. After we have those, we need to design an experiment whose outcome we can predict using the hypothesis being tested [Eq.

(15.10) in this case]. We will not go into detail about how to measure the value of the E!

-field and the -fieldV , but will instead focus on an experiment we will use to test the hypothesis. It involves the

interaction of the electric field with air molecules. We learned in Chapter 14 that the air (an electric insulator under normal conditions) could

turn into a conductor if free electrons, present in the air, are accelerated to such speeds that they can knock an electron out of an air molecule on the next collision (see Example 14.14). The two electrons then move on and knock additional electrons out of other air molecules creating a cascade, called dielectric breakdown. When these electrons recombine with molecules that have lost an electron, light is produced—a spark. Experiments have been done that determine that dielectric breakdown occurs in

dry air when the magnitude of the E!

-field exceeds about 63 10 V m, .

Now that we understand this part, we can conduct the testing experiment. For the experiment we use a Van de Graff generator that we learned about in Chapter 14. A Van de Graff generator with a sphere of 30 cm radius normally reaches a -fieldV value of 450 kV before the accumulated electric charge on the dome starts escaping into the air. Thus for this experiment we assume that the dome of a fully charged generator has a -fieldV of 450,000 V with respect to ground (which we

will consider to be at zero potential).

Testing Experiment Table 15.2 Testing the relationship between E!

field and -fieldV . Testing experiment Prediction Outcome

Use a Van de Graf generator and a metal sphere on a wooden handle.

Charge the Van de Graf generator to 450,000 V and do not charge the second sphere. The two spheres have a constant potential difference

= (450,000 V – 0) = 450,000 VV2 . Now move the sphere on the handle closer and closer to the generator sphere. Predict the distance between them when you see a spark.

Note that = (450,000 V – 0)V2 between the spheres. If Eq. (15.9) is correct, and we put the spheres 0.15 m apart, the magnitude of the E

!

field should be 6450,000 V

= = 3.0 10 V/m0.15 m

V

x

2,

2,

enough to cause breakdown. Thus, we should see a spark when the spheres are about 0.15 m = 15 cm apart.

We slowly approach the generator with the sphere. A spark is seen when the distance between them is about 15 cm.

Conclusion The outcome of the experiment matches the prediction. This does not prove that the potential difference V2 and the E

! field are related by Eq. (15.9) but does provide some experimental support for it.


Suppose you shuffle across a rug and then try to open a door using a metal doorknob. Just

before you touch the doorknob, a spark jumps between you and the knob. We estimate that the distance the spark jumped was about 1 cm. We can now use the value for the dielectric breakdown of air to estimate the potential difference between your body and the doorknob. A quick estimate yields:

) *) *6 43 10 V m 0.01 m 3 10 V=30,000 VV E x2 " 2 " , " ,

This is a huge potential difference. Fortunately, the potential difference is not dangerous; the amount of electric charge that flows is what is dangerous—and is very small in this case. Quantitative Exercise 15.10 Reducing air pollution with electric field An electrostatic precipitator

used to remove pollutants from a factory’s rectangular chimneystack has two metal plates on the sides of the stack (Fig. 15.24.) The metal plates are separated by 0.20 m and there is a 400-V potential difference between the plates. The purpose of the electric field is to exert an electric force on charged particles flowing up the stack, causing them to be accelerated toward and captured on the sidewalls instead of being released into the environment. What is the average

magnitude of the E!

-field between the plates?

Figure 15.24 Electrostatic precipitator

Represent Mathematically Use Eq. (15.11) to relate the magnitude of the E!

-field to the potential difference between and separation of the plates (absolute value symbols use to find the magnitude):

avVEs

2" +

2

Solve and Evaluate

400 V 2000 V/m0.2 mav

VEs

2" + " "

2

Such precipitators spray electrons (negatively charged) that attach themselves to the smoke and dust particles at the bottom of the chimney allowing the electric field to filter them out. Try It Yourself: Determine the horizontal acceleration of a dust particle going up this chimney. The

particle has mass 61 10 kg+, and charge 92 10 C+, .

Answer: 4 m/s2. Review question 15.7

Suppose the electric E!

-field is zero in some region of space. Is the V-field in this region also zero? Explain.


15.8 Conductors in E!

-fields

The concepts of the E!

-field and the V -field can be used to understand more precisely the behavior of electrically conducting materials. The situations can be separated into two general categories: 1) when a charged conductor is producing a contribution to the electric field, and 2) when a conductor is not charged but is placed in a region with an electric field produced by other objects. Let us consider these two cases separately. Electric field of a charged conductor

We already discussed in Section 15.3 that if you have a uniformly charged infinitely large

metal plate, it produces a uniform E!

-field which can be represented with E!

-field lines that are perpendicular to the surface of the plate and equally spaced. What if you have a spherically shaped conductor? Imagine that we have a metal sphere of radius R that we touch with a negatively charged plastic rod (Fig. 15.25a). Some of the excess electrons on the rod move to the metal sphere (Fig.

15.25b). The electrons transferred to the sphere create an E!

-field in the metal that causes free electrons to accelerate away from that spot. These excess free electrons inside quickly redistribute

until equilibrium is reached where the E!

-field inside the conductor becomes zero (Fig. 15.25c).

Figure 15.25 Charging a metal sphere

Since the E!

-field is zero within and on the surface of the sphere, this also means that the V -field has the same value at all points on the sphere. If there were an electric field within the sphere, then there would be two points inside with different values of the V -field and the electric charges would keep moving forever.


The electric field lines and equipotential surfaces are shown in Figs. 15.26a and b for a point-like charged object with charge +q and for the metal sphere with charge +q on its surface (which has radius R.)

Figure 15.26(a)(b) E field and constant V lines for point charge and charged sphere

Outside the metal sphere the fields look the same. We won’t do the calculation here since it requires

calculus, but the magnitude of the E!

-field and the value of the V -field outside the sphere are the same as that produced by a point-like charged object:

2 and q qE k V kr r

" " ,

where r is the distance from the center of the sphere ( r R7 ). Figures 15.26c and d show graphs of

E(r) and V (r) of the charged metal sphere. Inside the sphere 0E " and qV kR

" , the same value it

has on the surface.

Figure 15.26(c)(d)

Grounding This property of a conductor that all points on its surface are at constant V -field value has a

very important application – grounding. When you ground a conducting object, you run a wire from it to the ground. What is the purpose of this wire and why are electrical devices such as electronics and kitchen appliances dangerous if not grounded?

Imagine that we have two conducting metal spheres, one of radius 1R with charge 1q and the

other sphere of radius 2R with charge 2q (Fig. 15.27). If we connect them with a long metal wire,


electrons will move between and within the spheres until the V-field on the surfaces of and within

both spheres achieves the same value. Therefore, the E!

-field within the spheres and the wire will also be zero.

Figure 15.27 Two charged metal spheres connected by wire

What will be the electric charge of each sphere now? Since the spheres are separated by a

significant distance (hence the long wire) we can reasonably assume that the V -field near the surface of each sphere is contributed to only by that sphere. Therefore

11

1

kqVR

" and 22

2

kqVR

"

However, since 1 2V V" we set the right sides of the equations to be equal and simplify them to

1 1

2 2

q Rq R" .

This result tells us that the bigger sphere has a greater electric charge than the smaller sphere. When we ground a conducting object, we are effectively connecting it with a wire to a sphere

of radius is = 6400 km = 6,400,000 mR , the radius of the Earth. Because of this, the electric

charge of the conducting object ends up being extremely close to zero. In addition, it means the value of the V-field on the surface of the conducting object and on the surface of the Earth is the same. This means you can safely touch the conducting object (which might be a toaster or HDTV) while standing on the ground without risk that you will experience an electric shock. Quantitative exercise 15.11 Grounding Consider two metal spheres, where sphere 1 has a radius that is 10 times greater than sphere 2. Large sphere 1 is originally uncharged and small sphere 2 has charge q. What is the charge on each sphere after connected by a long conducting wire (Fig. 15.28)?

Figure 15.28 “Grounding” small sphere to large sphere


Represent Mathematically After connection, the V -field value on the surface of each sphere is the same:

1 21 2

1 2

1 1

2 2

kq kqV VR R

q Rq R

" " "

/ "

The total charge of both spheres must add to the initial charge q of sphere 2. Thus,

1 2q q q. " .

Solve and evaluate The first equation gives us:

1 1

2 2

110

q Rq R" "

Thus, q2 = 10q1. Therefore:

1 110q q q. " ,

or

1 11qq "

From these two equations we find that 10/11 of small sphere 2’s original charge q will be transferred to large sphere 1’s and 1/11 will remain on small sphere 2.

Try It Yourself: Suppose a Van de Graff generator of radius 0.10 m has a charge of about 61 10 C++ ,

or the charge of about 126 10, electrons. The Van de Graff generator is then turned off and grounded. Estimate how many excess electrons remain on its dome.

Answer: ) *17VdG 19

1 electron1.6 10 C 100 electrons1.6 10 C

q ++

# $" + , 8% &+ ,' (.

Uncharged conductor in an electric field

In the chapter opening we had a story about a person sitting in a car during a lightning storm. To explain why it is safe to be inside the car we need to learn about the behavior of uncharged conductors placed in regions where other charged objects have contributed to the electric field.

Imagine that we take a hollow conducting object and place it in a region with uniform nonzero E!

-

field whose value is 0E!

(Fig. 15.29a). Let’s analyze what happens along the left and right sides of the

object. The free electrons inside the object redistribute, producing their own contribution 1E!

to the

E!

-field (Fig. 15.29b.) This continues until the E!

-field within the conducting object is reduced to zero (Fig. 15.29c.):

0 1 0E E E" . "! ! !

.


The E!

-field produced by the environment 0E!

is cancelled by the contribution produced by the

conductor. The net E!

-field outside the conducting object is now a combination of the two E!

-fields

(Fig. 15.29d). Note that electrons inside the conductor have rearranged themselves until the E!

-field is perpendicular to the conducting surface and does not drive any further motion in the conductor.

Figure 15.29 A hollow conducting object in an external E0 field

Note that because the E!

-field inside the conductor is zero, a person inside a car during a lightning storm is safe. Note that the person shown in Fig. 15.30 inside a metal screen is unaffected by the electric discharge from the large Tesla coil. We can now understand what happened in Observational Experiment Table 15.1. When a metal can covers the electroscope, the can creates its

own E!

-field contribution that cancels the E!

-field contribution of the charged rod. This shielding

property of conductors is used to protect sensitive electronic devices from E!

-fields in their environment.

Figure 15.30 Shielded by metal screen

Review Question 15.8 Use the ideas of shielding to explain the experiments in the first section of this chapter.


15.9 Dielectric material in an electric field We learned in the previous section that conducting materials placed in an external electric field

produce their own contribution to the E!

-field that inside the conductor cancels the E!

-field produced by their environment. In this section we investigate the behavior of a dielectric (non-conducting) material placed in an external electric field. Recall from Chapter 14 that uncharged objects made of dielectric materials are attracted to both positively and negatively charged objects. This behavior was explained using the idea that the atoms in the dielectric material became polarized due to the presence of the charged object. All materials are made of atoms, which are comprised of positively charged nuclei surrounded by negatively charged electrons (Fig. 15.31a). If you place an atom in a region with an

0E!

external electric field, the field exerts a force on the atom’s positive nucleus in the direction of the

0E!

field and a force in the opposite direction on the atom’s negatively charged electrons. The nucleus

and the electrons are displaced slightly in opposite directions away from each other until the force exerted on each of them by the field is balanced by the force that they exert on each other due to their attraction (Fig. 15.31b). Now, instead of an atom with the center of its negative charge coinciding with the center of its positive charge, the center of the positive and negative charges are spatially separated. Such a system is said to have an electric dipole (Fig. 15.31c). One can represent an electric dipole using a vector that points from the negative charge center of the system to its positive charge center.

Figure 15.31 E field polarizes an atom

Some molecules, such as water, have electric dipoles even before they are affected by the E!

-field of their environment (Fig. 15.32a). The water molecule has four electron pairs with six electrons from oxygen and two from hydrogen atoms (there are also two other oxygen electrons deep inside the oxygen atom). These four electron pairs are distributed in four tetrahedral oriented electron clouds. Two of these electron pairs are each bonded with a positive proton (the nuclei of hydrogen atoms). The other two electron pairs do not have protons. Thus, there is a net positive electric charge on the


side with protons and a net negative charge on the side with the electron pairs with no protons. When

the external E!

-field is zero, water molecules are oriented randomly due to their bumping into each

other (Fig.15.32b). When the external E!

-field is nonzero, the electric field exerts forces in the opposite direction on the ends of the molecules producing a torque that tends to rotate the molecules

so that their electric dipoles align with the E!

-field (Fig. 15.32c).

(c) Figure 15.32

When a dielectric material is placed in a non-zero E!

-field, the atomic and molecular electric dipoles that result form a layer of positive electric charge on one surface of the material and a negative layer on the other surface (Fig. 15.33a). These layers produce an additional contribution to

the E!

-field, an 1E!

inside the dielectric material that points in the opposite direction to the external 0E!

field (Fig. 15.33b). The net electric field inside is 0 1E E E" .! ! !

. Since 1E!

points opposite 0E!

,

0E E6! !

. Thus, dielectric materials reduce the magnitude of the E!

-field inside the material, similar to

what happens in conductors. However, unlike conductors, the E!

-field within does not decrease to zero.

Figure 15.33 E0 field causes polarization and internal E1 field

The ability of a dielectric material to decrease the E!

-field varies from material to material. Physicists use a new physical quantity to characterize this ability - the dielectric constant 9 "Greek


letter kappa) of the material. It is a measure of how much the material reduces the external E!

-field (the field produced by the environment) within itself. The larger the value of 9 , the more the

reduction of the 0E!

-field by the dielectric material.

The dielectric constant of a material is defined as follows:

0EE

9 " (15.12)

where 0E is the magnitude of the E!

-field produced by the environment and 0 1E E E" + is the

magnitude of the E!

-field within the dielectric material. You can see from Eq. (15.12) that the dielectric constant is dimensionless. Sample dielectric constants for different types of materials are provided in Table 15.3. Using this definition, we can calculate the magnitude of the electric force that one charged object (object 1) exerts on another charged object (object 2) when both are inside a material with a dielectric constant 9

1 in vacuum 1 on 2 in vacuum1 on 2 in dielectric 2 1 in dielectric 2 E FF q E q

9 9" " "

The force that object 1 exerts on object 2 is reduced by 9 compared with the force it would exert in vacuum. Inside the dielectric material, Coulomb’s law is now written as:

1 21 on 2 2

q qF kr9

" (15.13)

Note in Fig. 15.34 that the positive ends of the water dipole molecules group around a negative ion and in effect reduce its net charge. The same thing happens to a positive ion in water. There is a reduction in the force of charged objects exert on each other when in that dielectric material. We can interpret Eq. (15.13) in this way:

1 2

1 on 2 2

q q

F kr9

# $% &' (" .

The charge product 1 2q q in Coulomb’s law has been reduced by a factor 19 .

Figure 15.34 Force between positive and negative ions reduced by polar water molecules

What is the value of the dielectric constant 9 for a conductor? Recall that when a conductor

is in the external electric field, the free charged particles move to produce their own electric field until


the E!

-field inside the conductor is zero. This means that the net force between two charges objects inside a conductor is always zero. In order for this to be consistent with Coulomb’s law, it means that for a conductor 9 is infinite! Sodium in blood The presence of dielectric materials (like water) has dramatic effects on processes occurring inside the human body. The body requires 5-10 grams of salt to function properly. The salt does not participate in any metabolic process. Instead the sodium ions from the dissolved salt are used in the transmission of information by the nervous system. Why in blood does solid salt dissolve into sodium

( Na. ) and chlorine ( Cl+ ) ions? Table salt comes in the form of an ionic crystal. Because a chlorine atom is more attractive to

electrons than a sodium atom, an electron is transferred from a sodium atom (leaving it with electric

charge 191.6 10 C+. , ) to the chlorine atom (giving it electric charge 191.6 10 C++ , ). The salt crystal is then held together by the electric force that the oppositely charged sodium and chlorine ions exert on each other. The distance between the two ions is about the same as the characteristic size of a

molecule (about 1010 m+ ). The magnitude of the electrical force that the ions exert on each other is then:

19 2Na Cl 9 2 2 -8

Na on Cl Cl on Na 2 -10 2

(1.6 10 C)(9 10 Nm /C ) 2 10 N(10 m)

q qF F k

r

+," " " , 8 ,

The electric potential energy of a sodium-chlorine ion system is about: 19 2

9 2 2 -18Na Cl-10

–(1.6 10 C) (9 10 Nm /C ) –2 10 J(10 m)q

q qU kr

+," " , 8 ,

The energy is negative because of the opposite charges of the ions. To separate the ions, approximately this much energy would need to be added to the system.

The ions also have positive kinetic energy due to their random thermal motion. A rough

estimate of this energy is 3 '2

K k T" , an expression we learned in Chapter 9 for the kinetic energy of

the particles in an ideal gas, where 'k is Boltzmann’s constant (do not confuse it with the k used in Coulomb’s law). This positive kinetic energy at a room temperature (300 K) is approximately:

) *23 -213 3' (1.4 10 J/K) 300 K 6 10 J2 2

K k T +" " , 8 ,

This positive kinetic energy of the individual ions is much smaller than the negative electrical potential energy holding them together. Thus the total energy of the system is negative and the ions remain bound together when salt is in air.

When the salt is placed in water or blood, two changes occur. First, compared with the salt crystal in air, there are many more collisions with the molecules in their environment. This means that although most of the collisions will not break an ion free from the crystal, some might. Remember,


3 '2

K k T" is the average kinetic energy of the molecules. A few molecules will have enough

random kinetic energy to knock an ion free from the crystal during a collision. Second, any ions that do become separated from the crystal by collisions do not exert nearly as strong an attractive force on each other because of the dielectric effect of the blood. Assuming the blood can be thought of as mostly comprised of water, the attractive force that the two ions exert on each other decreases by a factor of 81, the dielectric constant of water:

2 19 29 2 2 -10

Na on Cl in water 2 -10 2

(1.6 10 C) (9 10 Nm /C ) 3 10 N81(10 m)

eF kr9

+," " , 8 , .

The electric potential energy is also smaller by a factor of 81: 2 19 2

9 2 2 -20-10

–(1.6 10 C)(9 10 Nm /C ) –3 10 J81(10 m)q

eU kr9

++ ," " , 8 , .

This energy is almost the same as the average kinetic energy of the random motion of the water molecules; thus the total energy of the system becomes zero (sodium and chlorine ions can separate from each other!). This means that the random kinetic energy of the water molecules is sufficient to keep the sodium and chlorine ions from recombining. This allows the nervous system to use this freed sodium ions to transmit information. Table 15.3 Dielectric Constants for Different Types of Materials Type of Material Dielectric Constant (9 )

Vacuum 1.0000 Dry air 1.0006 Wax 2.25 Mica 2 – 7 Glass 4 – 7 Benzene 2.28 Paper 3.5 Axon membrane 8 Body tissue 8 Ethanol 26 Methanol 31 Water 81 * At 20 oC.

Review Question 15.9 What is the difference between conducting and dielectric materials when they are placed in a region

with nonzero E!

-field? Explain these differences.

15. 10 Capacitors and Electric Field Energy We learned about practical applications of conductors in electric fields, such as grounding

and shielding. Another important application involving electric fields and conductors concerns the


storage of energy in the form of electric potential energy. When positively and negatively charged objects are separated, the system possesses electric potential energy. How can this charge separation be maintained so that the electric potential energy can be stored for useful purposes? This is accomplished with a device known as a capacitor. Capacitors

A capacitor consists of two conducting surfaces separated by a non-conducting material. Although a variety of configurations are possible, the simplest are parallel plate capacitors comprised of two metal plates separated by air, rubber, paper, or some other dielectric material (Fig. 15.35a.) When the conducting plates are connected with metal wires to the terminals of a battery, the plates become charged. As the electric charge is a conserved quantity, the appearance of those opposite charges can be explained if we assume that the battery somehow removes negative electrons from one plate and deposits them to the other. The plate with a deficiency of electrons is positively charged with charge q. and the plate with excess electrons is negatively charged with change –q (Fig.

15.35b).

Figure 15.35(a)(b) A capacitor

If we consider the capacitor plates to be large flat conductors, the charges should distribute

evenly on the plates. Since each charged plate produces a uniform E!

-field, the sum of the

contributions from each plate is also a uniform E!

-field between the plates that is twice as strong as

that from one plate (see Fig. 15.35c). Outside the plates, the E!

-field from the positively charged

plate points opposite the E!

-field from the negatively charged plate. Thus, they cancel resulting in a

zero E!

-field outside the plates (Fig. 13.35d).

Figure 15.35(c)(d)


According to Eq. (15.11), the magnitude of the E!

-field between the plates depends on the potential difference V2 (difference in V-field values) from one plate to the other and on the distance d separating them:

V VEx d

2 2" + "

2

Thus, if the potential difference across the plates doubles (for example, you connect the capacitor to a

12 V battery instead of 6 V ), the magnitude of the E!

-field between the plates also doubles. Recall

that each E!

field lines originate on positive charges and ends on negative charges. Thus, to double

the E!

field the charge on the plates has to double. We conclude that the magnitude of the charge q on the plates ( q. on one plate and q+ on the other) should be proportional to the potential difference

#V across the plates: q V: 2 . Or if we use a proportionality constant,

q C V" 2 . (15.14)

The proportionality constant C in this equation is called the capacitance of the capacitor. The absolute value is needed since q in this context is the positive magnitude of the charge on each plate. The unit of capacitance is 1 coulomb/volt = 1 farad (1 C/V = 1 F) in honor of Michael Faraday whose experiments helped establish the atomic nature of electric charge. Note that a large capacitance capacitor will maintain more charge ( q. and q+ ) on its plates than a capacitor with small

capacitance, even if both have the same potential difference from one plate to the other. Capacitance of a capacitor

What properties of capacitors determine their capacitance? It might seem that Eq. (15.14) answers this question. However, the capacitance of a capacitor does not change when the charge on the plates or the potential difference across them is changed. These two quantities are proportional to each other ( q V: 2 ); to double the charge on the plates you need to double the potential difference

across the same capacitor. But the ratio C q V" 2 remains the same. This ratio is an operational

definition of capacitance. What actually determines the capacitance of a particular capacitor? Imagine two capacitors

whose plates have different surface areas and are connected to the same potential difference source. The capacitor with the larger surface area A plates should be able to maintain more charge separation as there is more room for the charge to spread out (Fig. 15.36a).

Figure 15.36(a) Capacitance depends on A, d and x


Now imagine two capacitors with the same surface area and the same potential difference

across the plates but with different distances d between the plates. Since V Ed2 " and V2 is

constant, a larger the distance d between the plates leads to a smaller magnitude E!

-field between

the plates. But, since this E!

-field is proportional to the amount of electric charge on the plates, a larger plate separation leads to a smaller magnitude of electric charge on the plates (+ on one and – on the other). This means the capacitance of the capacitor decreases with increasing d (Fig. 15.36b.)

Figure 15.36(b)

A third structural quantity that affects the capacitance is the dielectric constant 9 of the

material between the plates. Material between the plates with a large dielectric constant becomes polarized by the electric field between the plates. The positive ends of the polar molecules form a layer of positive charge near the negatively charged plate and the negative ends of the polar molecules form a layer of negative charge near the positively charged plate (Fig.15.36c). The polar molecules near the plates tend to cancel the effect of some of the charge on the plate and it easier for more charge to move onto the plates. Thus, more charge moves onto the capacitor plates for capacitors whose plates are separated by material of high dielectric constant. The capacitance increases in proportion to the dielectric constant 9 of the material.

Figure 15.36(c)

Thus we conclude that the capacitance of a particular capacitor should increase if the surface

area A of the plates increases, decrease if the distance d between them is increased, and increase if the dielectric constant 9 of the material between them increases. A careful derivation (which we will not go through) provides the following result for a parallel plate capacitor:

Parallel plate capacitor 4ACkd

9;

" (15.15)

where 9 2 29.0 10 N m Ck " , - .


Tip! There are two similar looking symbols in Eq. (15.15). 9 is the dielectric constant of the material between the plates, and k is the constant in Coulomb’s law. Quantitative Exercise 15.12 Capacitance of Textbook First, estimate the capacitance of your

physics textbook if the front and back covers (area 20.050 mA " , separation 0.040 md " ) were made of a conducting material. The dielectric constant of paper is approximately 6.0. Second, determine what potential difference must be produced across its covers in order for the textbook to

have a charge separation of 10-6 C (one plate has charge 610 C+. and the other has charge 610 C++ ). Represent Mathematically The capacitance of a parallel plate capacitor is:

4ACkd

9;

"

The potential difference V needed to produce a charge separation of 610 Cq +" is:

qVC

2 "

Solve and Evaluate

) *) *) *) *

211

9 2 2

64

11

6.0 0.050 m7.0 10 F 70 pF

4 4 9.0 10 N m C 0.040 m

10 C 1.4 10 V 14 kV7.0 10 F

ACkd

qVC

9; ;

+

+

+

" " " , ", -

2 " " " , ",

Let’s check the units for capacitance:

) *2 2 2

2

2

m C C C CC FN m J J VN m m

C

" " " " "-# $-

% &' (

The units check out. This is a rather small capacitance, but reasonable because the plate separation is quite large and the dielectric is not so good. The farad is a very large unit though, and capacitors with picofarad or nanofarad capacitances are quite common. Try It Yourself: How big approximately should a book be to have a capacitance of 1 F? Answer: If we assume the same plate area and separation, then the covers should be about 106 m long or about a thousand kilometers! This is an illustration of just how large a unit the farad is. Body Cells as Capacitors Capacitors have numerous applications. Capacitors can be found inside camera flashes, the tuning circuits of radios, and music amplifiers. Biological capacitors are found inside our bodies. Cells, including nerve cells, have capacitor-like properties (see Fig. 15.37). The conducting “plates” are the fluids on either side of a moderately non-conducting cell membrane. In this membrane, chemical processes cause ions to be “pumped” across the membrane. As a result, the membrane’s


inner surface becomes slightly negatively charged while the outer surface becomes slightly positively charged.

Figure 15.37 Body cell is a capacitor

Example 15.13 Capacitance of and charge on body cells Estimate (a) the capacitance C of a single cell and (b) the charge separation q of all the membranes of a person’s 1013 body cells. Assume that

each cell has a surface area of 9 21.8 10 mA +" , , a membrane thickness of 98.0 10 md +" , , a 0.070-VV2 " potential difference across the membrane wall, and a membrane dielectric constant

8.09 " . Sketch and Translate A body cell is sketched in Fig. 15.37. The information needed to answer the questions is given in the problem statement.

Simplify and Diagram The thickness of the membrane 98.0 10 md +" , is much less then the

dimensions of the cell (roughly the square root of the surface area 54.2 10 mA +" , .) Thus, when very close to the cell membrane, it looks almost flat, in the same way Earth appears flat when you are close to its surface. Thus, we can use the expression for the capacitance of a parallel plate capacitor to make our estimate. Represent Mathematically The capacitance of one cell is:

4ACkd

9;

" .

The total charge separation q on the 1310 cells (biological capacitors) in the body is then

) *13total 10q C V" 2

Solve and Evaluate The capacitance of one cell is:

) *) *

) *) *9 2

119 2 2 9

8.0 1.8 10 m1.6 10 F

4 9.0 10 N m C 8.0 10 mC

;

++

+

," " ,

, - ,

The total capacitance of all 1310 cells is then: 13 –11

total 10 (1.6 x 10 F) = 160 FC " .

The total charge separated by the membranes of all of the cells is approximately:

total = (160 F)/(0.070 V) = 11 C.q C V" 2


Although these calculations are approximate, it is clear that the separation of electric charge (-11 C total on the inside walls of cell membranes and +11 C on the outside walls) is an important part of our metabolic processes. This electric charge separation 11 C is huge—about the same as the charge transferred during a lightning flash. Fortunately, our bodies’ cells do not all discharge simultaneously in one surge. Try It Yourself: Suppose you doubled the wall thickness of all body cells. How would the potential difference across them have to change in order for the cells to maintain the same charge separation? Answer: Doubling the wall thicknesses would reduce the capacitance by half. Thus, you would need to double the potential difference across the walls to maintain the same charge separation.

Energy of a charged capacitor

We learned in Chapter 14, that a system of positively and negatively charged objects that have been separated from each other has electric potential energy. This is precisely what a capacitor is. How much electric potential energy does the system comprised of the two oppositely charged plates of a parallel plate capacitor have? To answer this question we start with an uncharged capacitor, then calculate the amount of work that must be done on the system to move electrons from one plate to the other (Fig. 15.38a.) More precisely, we move increments of charge – q2 (negative

since these charge increments are comprised of electrons) from one plate to the other. After the first charge increment is moved, the plate from which it was taken has a charge of q.2 and the plate to

which it is taken now has a charge of – q2 .

Figure 15.38 Charging a capacitor

Only a very small amount of work must be done to move this first – q2 since the left plate is

uncharged (Fig. 15.38b.) The next – q2 is more difficult to move (Fig. 15.38c) since the left plate

now has a charge of – q2 which repels the – q2 we are trying to move there. Additionally, the

2 q. 2 charge on the right plate pulls back on it. The more charge increments we move from one


plate to the other, the more difficult it becomes to move the next one. Eventually, with increasing

effort, we will have transferred a total (negative) charge of ) * ) *–q q q" +2 . +2 ." to the left plate

leaving the right plate with a charge of +q. We can represent the whole process of charging the capacitor with a work-energy bar chart (Fig. 15.39). The external object (us, moving the charge increments) does work on the system (the capacitor) and the electric potential energy of the system increases.

Figure 15.39 Work done to charge a capacitor

The discussion above was a thought experiment. It’s not possible for us to manually grab electrons and move them from one capacitor plate to the other. Usually, this external source that does this work is a battery. The electrical potential energy increase can be calculated using Eq. (15.7):

averageqU q U2 " 2 ,

where averageV2 is the average potential difference from one plate to the other during the process of

charging. Since the initial potential difference is zero and the final is V2 , the average potential difference between the plates is:

average0

2 2V VV . 2 2

2 " " .

Substituting this expression for averageV2 into Eq. (15.7), we get:

average 2qVU q V q 2" 2 " .

The above expression for the electric potential energy of a charged capacitor can be written in three

different ways using Eq. (15.14), q C V" 2 :

221 1 1

2 2 2qqU q V C VC

" 2 " 2 " . (15.16)

Quantitative Exercise 15.14 Energy needed to charge human body cells In Example 15.15, we estimated that the total charge separated across all the cell membranes in a human body was about 11 C. Recall that the potential difference across the cell membranes was 0.070 V. Estimate the work that

must be done to separate the charges across the membranes of these approximately 1310 cells. The situation for one body cell is shown in Fig. 15.37. Represent Mathematically We can use the first expression in Eq. (15.16) to answer this question:


12qU q V" 2

where 11 Cq " and 0.070 VV2 " .

Solve and Evaluate

) *) *1 1 11 C 0.070 V 0.40 J2 2qU q V" 2 " "

These cells are continually being charged and discharged as part of the body’s metabolic processes. Try It Yourself: Estimate how much energy is needed to charge the body cells each day assuming that each cell discharges once per second. Answer: 35,000 J = 8 kcal or about one-tenth the energy provided by a piece of bread. This is not much. However, some cells might discharge up to 400 times per second, though some might discharge fairly infrequently, so this result is just a rough estimate. History of capacitors

The capacitor was invented almost simultaneously by two people: in 1745 by the German jurist, Lutheran cleric, and physicist Ewald Georg von Kleist and in 1746 by the Dutch physician and physicist Pieter van Musschenbroek. They were interested in making a device that could maintain a separation of positive and negative charges. At that time, charge separation could be produced by friction machines (rubbing, similar to Whimhurst generators), but there was no way to maintain it. Musschenbroek and his student Andreas Cunaeus discovered that the charge separation could be maintained in what became the first capacitors. They called these devices Leyden jars (they worked at

the University of Leyden). A simple Leyden jar is literally a jar made of a dielectric

material (glass originally) lined with a conducting material both inside and outside the jar (Fig. 15.40). The outside conducting material is connected with a wire into the ground. To charge this capacitor, the inside conducting material is connected to the friction machine. Suppose this causes the inner conductor to become negatively charged; then the outer conductor will become positively charged as electrons on the outer conductor are repelled into the ground. Thus, the inner and outer surfaces of the jar serve as capacitor plates. Since there is a dielectric material between them, the charge separation is maintained.

Figure 15.40 A Leyden jar capacitor

You can test that a capacitor is charged by disconnecting the grounding wire from the ground and bringing that wire close to the inner conductor. You will see a spark between the wire and the inner conductor. Experiments by the inventors of the Leyden jar revealed that the amount of charge stored by a capacitor increased with increased conducting surface areas and decreased thickness of


the dielectric container walls. A typical Leiden jar has a very small capacitance – between 1250 10 F+, and 910 F+ . Contemporary capacitors used routinely in electric circuits have

capacitances on the order of 10-6 F, but 1 F capacitors are now available inexpensively for physics laboratory experiments. Energy density of electric field

As we discussed above, the capacitors were invented as devices whose goal was to maintain charge separation, and therefore store electric potential energy. We found that the energy stored in a

charged capacitor is 212qU C V" 2 . Where is this energy stored? One way to answer this is to

notice that the difference between a charged and an uncharged capacitor is the presence of the electric field in the region between the plates. Perhaps that is where the electric potential energy is stored. This is a very interesting suggestion, since up until now we have thought of the electric field only as a mechanism to explain how charged objects can interact without being in physical contact. Suggesting that the electric field has electric potential energy makes the electric field seem much more real than just an idea.

Since the electric field fills the region between the plates, we can describe the energy that it has in terms of an energy density, much in the same way we can characterize an object that fills a region as having a mass density. This energy density quantifies the electric potential energy stored in

the electric field per cubic meter of volume. Quantitatively, the E!

-field energy density Eu is:

qE

Uu

V"

where qU is the electric potential energy stored in the E!

-field in that region, and V is the volume of

the region.

Tip! Notice that here the letter V stands for the physical quantity of volume, not for the electric potential. Sometimes similar or identical symbols are used for different physical quantities. When looking at a mathematical expression, always ask yourself: “what do each of these symbols represent?”

The above equation for the electric field energy density is not written in terms of the E!

-field. Let us rewrite it using our knowledge of the electric energy of a parallel plate capacitor:

) *22 11

2 42qE

A EdC VU kduV V V

9;

# $# $# $2 % &% &% & ' (' ( ' (" " "


We used Eqs. (15.15) and (15.10) to substitute for C and V2 . A is the plate area, d is the plate

separation, E is the magnitude of the E!

-field between the plates, and 9 is the dielectric constant of the material between the plates. Simplifying this we have:

) *2

2

8 8E

A Ed Adu EkdV k V

9 9; ;

" "

Since the volume of the region between the plates is V Ad" this equation simplifies to:

2

8Eu Ek

9;

" (15.17)

The energy density depends only on the magnitude of the E!

field, the properties of the dielectric, and a few mathematical and physical constants.

Let’s check the units. Remember that the unit for k is 2 2N m C- , 9 has no units, and the

units for E are N/C. Therefore we get (note also that 1 N•m = 1 J): 2 2

2 2 2 3

C N N JN m C m m

" "�

.

The units are in fact the units of energy per volume, the correct unit for energy density.

It is not surprising that the energy density depends on the magnitude of the E!

-field, but why is it proportional to the dielectric constant of the material? Remember that the presence of the dielectric allows the plates of the capacitor to have greater positive and negative charges even though they are connected to the same battery (Fig. 15.41). This means more work needs to be done on the capacitor system to charge it, which means greater electric potential energy and consequently greater

energy density of the E!

-field.

Figure 15.41 More charge, E, and energy density with dielectric material

Tip! Although we derived the expression for energy density for a specific case of a parallel plate

capacitor, it can be applied for any E!

-field. However, since in general the E!

-field changes from

point to point, the value of the E!

-field energy density will change from point to point as well.


Review Question 15.10 Describe a method of charging a capacitor that will result in the two plates having equal magnitude but opposite sign charges.

15.11 Putting it all together: Electrocardiography and Lightning Understanding electric phenomena occurring in nature and in our technology can be made

much easier by using the ideas we have developed about the electric field. In particular, they will be extremely valuable in understanding electric circuits—the subject of the next chapter. To close this

chapter we use the E!

and V -fields to understand electrocardiography and lightning. Electrocardiography

The heart has four chambers (Fig. 15.42) that each pump blood in a special sequence. The upper right chamber (the right atrium) collects blood from a large vein, the vena cava. This blood is returning to the heart after a trip around the circulatory system. Before returning to the heart, this blood delivered the oxygen it was carrying to body cells and collected carbon dioxide and other waste products from them. After the blood’s return to the right atrium, the atrium pumps this blood with its

waste products into the right ventricle—a relatively large and muscular chamber that then pumps blood into the lungs. Here carbon dioxide is exchanged for fresh oxygen and returned to the upper left chamber of the heart, the left atrium. The left atrium pumps the blood into the biggest and strongest pump of the four chambers—the left ventricle. The left ventricle then compresses and sends the blood into the large artery known as the aorta for another trip around the circulatory system to provide the oxygen needed by the body cells.

Figure 15.42 The heart and lungs

This pumping sequence is repeated about once each second. The operation of the four chambers is closely synchronized. An electrocardiogram (ECG) is an electronic method to monitor this pumping cycle to detect heart problems—for example an enlarged left ventricle or an enlarged right ventricle. An ECG consists of several pads placed on the skin of the person which are connected to a device which interprets the data recorded by the pads. What are the pads measuring, and how can they determine anything about what is going on inside the heart?

An important part of this process involves the electric charge separation that occurs when muscle cells in the heart contract during the pumping process. For example, when the left ventricle pumps blood into the aorta, many muscle cells are contracting. As each muscle cell contracts, positive and negative charges separate as represented schematically in Fig. 15.43. The simultaneous contraction of the large number of cells in the muscle results in a relatively large charge separation—an electric dipole. On the other hand, the number of simultaneously contracting muscle cells is much


smaller when the left atrium pumps blood into the left ventricle. Thus, during the 1-s cycle of a heartbeat, there is a changing electric dipole that depends at each instant on the number and the orientation of the muscle cells that are contracting at that instant.

Figure 15.43 Electric dipole produced by muscle cell contraction

This electric dipole produces a V -field that extends outside the body, which the pads detect. The device connected to the pads determines the potential difference between the various pairs of pads, and then uses this data to reconstruct the changing electric dipole of the heart. Consider Fig. 15.44a, which shows a simplified electric dipole charge distribution of a heart at one instant of time and also shows two ECG pads on opposite shoulders of a person’s body. What do we expect these pads to measure at that particular instant? Consider the next exercise.

Conceptual Exercise 15.15 Heart’s Electric Dipole and ECG Potential Difference (a) Draw E!

-field vectors representing the electric field at the location of the dot in Fig. 15.44a.produced by the heart at that instant. (b) Decide the direction that the electric field exerts forces on a positive ion (such as Na+) and on a negative ion (such as Cl–) in the body tissue at that location. Sketch and Translate The situation is sketched in Fig. 15.44a. Each charge of the heart’s electric

dipole produces a contribution to the E!

-field at each point in space, including the location of interest.

To find the E!

-field at that point, we graphically add the E!

-field contributions produced by each dipole charge (see the Electric Field Reasoning Skill on p. xxx). The electric field at the location of

interest exerts a force on ions located there. The force points in the same direction as the E!

-field for positive ions and in the opposite direction for negative ions.

Figure 15.44(a) Heart’s dipole produces VI > VII


Simplify and Diagram The E!

-field contribution from each charge is shown in Fig. 15.44b along

with the graphical addition of these two contributions to estimate the direction of the E!

-field. The

E!

-field exerts a force on positive sodium ions toward ECG pad I and on negative chlorine ions toward ECQ pad II (Fig. 15.44c). This means the region near pad I will tend to be slightly positively charged while the region near pad II will tend to be slightly negatively charged (Fig. 15.44d.) This means pad I will measure a slightly higher V-field than pad II. The device the pads are connected to then detects this difference and uses it to infer the health of the heart.

Figure 15.44(b)(c)(d)

Try It Yourself: What change(s) in the situation in the last example would cause the potential difference between leads I and II to be qualitatively less than in the example? Answer: If the dipole charge on the heart was less (fewer muscle cells contracting) or if the orientation of the dipole was different (for example, the dipole was oriented more vertically).

We now see how the heart’s electric dipole causes a potential difference between different positions on a person’s body. When reading an electrocardiogram, the analysis process is reversed. The technician looks at the varying potential difference between different pairs of electrodes and from this can learn about the heart’s changing electric dipole. This dipole in turn indicates the number of muscle cells contracting at different times. A person with left ventricular hypertrophy (an enlarged left ventricle) will have an especially large dipole and potential difference during the contraction of


the left ventricle. Figure 15.45 shows a normal ECG and another produced by a person with left ventricular hypertrophy.

Figure 15.45 ECG for normal left ventricular heart beats

Dielectric Breakdown

We already learned about dielectric breakdown in Section 15.3. Here we return to this

question in more detail. If the E!

-field in air or in some other material is very large, free electrons in the air or material accelerate, quickly acquiring enough kinetic energy between collisions with other atoms and molecules to ionize them when collisions do occur. This produces additional free electrons which in turn accelerate, collide with, and ionize yet more atoms and molecules. This results in a very

intense cascade of electrons flowing opposite the E!

-field direction. This phenomenon, called dielectric breakdown, occurs suddenly and often produces a spark.

The dielectric strength of a material is defined as the magnitude of the E!

-field for which breakdown of the material occurs. Non-conducting materials—plastic, rubber, paraffin, and

transformer oils—have dielectric strengths in the range of 68 10, to 620 10 V m, . If the dielectric

strength of a material were 68 10 V m, , then the potential difference across a 1-m-thick slab of the

material would have to be 68 10 V, or greater for breakdown to occur. If the slab were 0.1 m thick,

then a potential difference of 58 10 V, or greater would cause breakdown. When you shuffle across a rug, charge separation occurs between your body and the rug (with

perhaps your body becoming positively charged.) If your hand is close to a conducting surface such as a doorknob, dielectric breakdown of the air can occur and a spark appears. Sparks in the air caused by dielectric breakdown have ignited explosions in grain elevators due to the highly flammable dust from the grain. Similar explosions can occur in operating rooms of hospitals if flammable anesthetic vapors are present. To prevent such explosions, physicians and nurses wear shoes with conducting soles. This prevents the charge separation and the resulting sparks. Quantitative Exercise 15.16 Potential difference during a doorknob spark Estimate the potential difference between your finger and a doorknob if a spark jumps between your finger and the


doorknob after you have walked across a carpet (Fig. 15.46.) The dielectric strength of air is 63 10 V m, .

Figure 15.46 Spark just before touching door knob

Represent Mathematically The relationship between the E!

-field magnitude in a region and the

potential difference across that region along the E!

-field direction is:

V Ed2 "

Solve and Evaluate The spark length is not very long—perhaps 31 mm 10 m+" Thus,

) *) *6 3breakdown breakdown 3 10 V m 10 m 3000 VV E d +2 " " , "

It is difficult to evaluate this estimate. This sounds like a rather large number. What is important is that it is not the potential difference across a region that is actually dangerous. What is harmful is when a large amount of electric charge flows through the body. This can disrupt the body’s metabolism (which we have learned is partly electrical in nature), or nervous system/brain function, or even stop the heart. Doorknob sparks involve little electric charge transfer and therefore aren’t dangerous. Perhaps your professor will demonstrate much larger sparks by holding the knuckle of his or her hand several centimeters from a Van de Graff generator—not something that would be considered fun by everyone. Try It Yourself: What is the potential difference between the Van de Graff and the professor if the spark jumps 5 cm? Answer: 150,000 V. Breakdown across body cell membranes

You might wonder about dielectric breakdown across the membranes of body cells. Recall from Example 15.14 that the membrane is about 8 nm thick and a normal potential difference across

it is 70 mV. Thus the magnitude of the E!

-field produced by the charge separation across the cell membrane is approximately

79

0.070 V 10 V m8.0 10 m

VE

d +

2" " 8

,


If our cell membranes were made of a material with dielectric strength less than 107 V/m, dielectric breakdown would occur. Fortunately, the dielectric strength of our cell membranes is about twice this value. Lightning

Another important application of dielectric breakdown is lightning. Perhaps the most common type of lightning (there are many types) is cloud-to-ground lightning. Lightning is a complex not completely understood phenomena with many different types. A simplified explanation of the lightning is depicted in Fig. 15.47. In (a) a cloud has become positively charged at the top (the P region) and negatively charged at the bottom (the N region). The reasons for this charge distribution are complex and seem to involve the rubbing of water droplets and ice particles in the cloud and water droplets of ascending air. The rubbing causes some particles to become negatively charged and others positively charged. The positively charged particles are less dense and accumulate at the top of the cloud while the denser negatively charged particles remain at the bottom. Electrons in the Earth directly below the cloud are repelled from the cloud and move away leaving the Earth positively charged under the cloud.

Figure 15.47(a) Lightning

If the magnitude of the E!

field caused by this charge distribution in the cloud is large enough, dielectric breakdown occurs, and electrons leap from the N region downward and away from the rest of the negative charge in the cloud (Fig. 15.47b) in a 15-30 m long stepped leader. The electrons make successive 15-30-m jumps toward the Earth at intervals of one-millionth of a second. This downward increasing length stepped leader (Fig. 15.47c) leaves a trail of negative ions. The eye does not see this stepped leader.

As the electrons approach the earth (Fig. 15.53d), the E!

-field in the air above a high point on the surface becomes so intense that positive ions from the earth (called the positive streamer) rush up to meet the stepped leader. A severe dielectric breakdown of air occurs. After this breakdown, negative charge in the stepped leader farther above the Earth can now rush down through the region of ionized air (Fig. 15.53e). This intense flow of electrons originating farther above the earth causes a flash of light that, when photographed, appears to move upward because electron flow starts from


successively higher and higher positions. The flash, which appears to move up, and the large downward electron flow make up what is called the return stroke. Eventually, a large number of electrons originally in the cloud will have made their way to the ground. About -5 coulombs worth of electrons move down during the stepped leader, and another -20 to -30 coulombs worth of electrons move down during the return stroke.

Figure 15.47(b)(c)(d)(e)

In eighteenth-century Europe, church bells were often rung to protect against lightning. Some

persons thought, erroneously, that the sound of the bells disturbed the path of the lightning and protected anyone near the bells. Unfortunately, the metal bells were usually at high elevations and thus were likely spots for lightning to strike. In one 33-year period during the eighteenth century, 386 bell towers were struck, and 103 bell ringers were killed at their ropes.

What should you do if you are caught outside during a lightning storm? You should not make a lightning rod of yourself or stand beneath something that will act like one (such as a tree). If you are in an open space, find a ravine, valley, or depression in the ground. Crouch so that you do not project above the surrounding landscape. Lightning usually strikes objects farthest above the Earth's surface. Do not lie on the ground. A large potential difference may develop between your head and feet, causing an undesirable flow of electric charge through your body. When crouched, keep your feet close together. The safest place during a storm is inside your car or other object made from a

conducting material. As discussed earlier, inside such an enclosure the E!

-field contribution of the

environment is cancelled by the E!

-field contribution of the enclosure.


Review Question 15.11 How do ECG pads attached to a person’s shoulders let doctors learn about the heart deep inside the chest of a patient?

Summary Word Representation Picture and

Diagram Representation

Mathematical Representation

E!

-Field is a physical quantity characterizing electric field at a point. The E

!-field at a point is determined

by imagining a pointlike positively charged object

Oq (the system) located at that point. The E

!-field

equals the net electric force exerted by other charged objects on

Oq divided by that object’s charge. The

units of E!

-field are newtons/coulomb (N/C). The E!

-field exerts a force on a charged object in the field.

O

O

other on Q qFE

q"

!!

(15.2)

Field on O O= F q E

##! ##! (15.4)

Superposition principle When n charged objects contribute to the E

!-field at a point, the E

!-field is

the vector sum of those contributions.

1 2 3E E E E" . . .! ! ! !

" (15.5)

V-field is another physical quantity characterizing electric field at a point. To determine its value, imagine a positively charged pointlike object oq at a position of interest. The V -field (electric potential) at that position is the electric potential energy qU of that object and all other charged charges divided by the charge of that object. Include the signs of charges when calculating qU . The unit of electric potential is the volt (V) where 1 V = 1 J/C (joule/coulomb). If know V at a point, then can find

qU at that point.

O

qUV

q" (15.6)

where

1 o 2 o

1 to O 2 to O

+ + ...qUkQ q kQ q

r r"

q qVU " (15.7)

Superposition principle for V -field When n charged objects contribute to the V -field at a point, the V -field is the vector sum of those contributions.

1 2 3V V V V" . . ." (15.9)

E!

-field and V field are related The greater the magnitude of the E

!-field at a particular location, the

faster the V -field changes with position near that location.

!V = –Ex !x (15.10) or

2 11 to 2

2 1

––

–xV V

Ex x

" (15.11)

Representing the E!

field and the V field E!

-field lines that start on positive charges and end on negative charges represent the E

!-field. Lines on the


surface with each point at the same V -field represent the V field. E

!-field lines point in the

direction of decreasing V -field.

Dielectric constant 9 A dielectric material partially cancels E

!-field within it produced by the

environment. This reduces the magnitude of the force exerted by charged objects on each other within the material

0E

E9 " (15.12)

1 21 on 2 2

q qF k

r9" (15.13)

Capacitors A capacitor is a device that (when charged) stores electric potential energy. It has two conducting surfaces separated by a non-conducting material. To charge a capacitor one connects its plates to the points at different electric potential (terminals of a battery for example). The charge separation results in the capacitor storing electric potential energy.

q C V" 2 (15.14)

4C

Akd

9

;" for parallel plate

capacitors (15.15) 21 1

2 2qU q V C V" 2 " 2 "

(15.16)

Conceptual Questions 1. Electric field is a (a) a law of physics; (b) a physical quantity; (c) a mental model; (d) a unit of

measurement. 2. A point-like charged object is surrounded by an electric field. The E

!field at point A which is a distance d

from the source charge: (a) depends on the magnitude of the test charge; (b) depends on the sign of the test charge; (c) depends on the magnitude and the sign of the source charge; (d) is independent of the test charge.

3. You can shield an object from an external electric field using a conductor because: (a) there are freely

moving electric charges inside conductors; (b) there are positive and negative charges inside materials; (c) both a and b are important.

4. If you place an object made of a conducting material in an external electric field E

!, the magnitude of the

E!

-field inside the material will be: (a) less than outside; (b) more than outside; (c) zero; (d) it depends on the size of the material.

5. Two identical positive charges are located at a distance d from each other. Where is the location at which

both the E!

field and the V field are zero? a) exactly between the charges; b) at a distance d from both charges; c) both a) & b) are correct; d) none of these choices.

6. How do we use the model of the electric field to explain the interaction between charges? 7. Describe a procedure that one can use to determine the E

! field at some point.

8. What does it mean if the E

! field at a certain point is 5 N/C and points north?


9. A very small positive charge is placed at one point in space. There is no electric force exerted on it. (a) What is value of the E

! field at that point? (b) Does this mean there are no electric charges nearby?

Explain. (c) Think of at least one charge distribution (two or more other charges) that would produce a zero E!

field at a point. 10. How do we create an electric field E

! with parallel lines and of uniform density (equally spaced)?

11. Draw a sketch of the E

! field lines caused by: a) two positive charges of the same magnitude; b) two

negative charges of the same magnitude; c) two positive charges of different magnitudes; d) a positive and a negative charge of the same magnitude; e) a positive and a negative charge of different magnitudes, and f) a positive charge between two negative charges. Explain in general how you decided on the direction of the lines.

12. Jim thinks that E

! field lines show the paths that test charges would follow if placed in the electric field.

Do you agree or disagree with his statement? If you disagree, give a counter example. 13. Can E

! field lines cross? Explain why or why not.

14. An electron moving horizontally from left to right across the page flies into a uniform E

! field that points

toward the top of the page. Draw vectors indicating the direction of the electric force exerted by the field on the electron and its acceleration. Draw a sketch indicating the path of the electron in this field.

15. Why do we need the V field as a quantity to describe the field produced by electric charges? Isn’t the E

!

field quantity sufficient? 16. (a) What does it mean if the V field at a certain point in space is 10 V? (b) What does it mean if the

potential difference between two points is 10 V? 17. You can say that the electric potential at a certain point is the electrical potential energy of a unit test

charge placed there. Is there any other way to determine the potential at a point? (Hint: Use the idea of work).

18. Explain how grounding works. 19. Explain how shielding works. 20. Explain the difference between the microscopic structures of polar and of non-polar dielectric materials.

Give an example of each. 21. Explain why we divide the force exerted by one charged object on the other by the value of the dielectric

constant in Coulomb’s law if the charged objects are submerged in a dielectric. 22. What does it mean if the dielectric constant 9 of water is 81? 23. Describe the relation between the quantities E

! field and V field.

24. Describe the relationship between the direction of E

! field lines and equipotential surfaces. Give examples.

25. If the V field in a region is constant, what is the E

! field in this region? Explain your answer.


26. A bird can stand on a high voltage power line with no negative effects (see Fig. Q15.26). How can this be?

Figure Q15.26

27. Why are uncharged pieces of a dielectric material attracted to charged objects? 28. Draw equipotential surfaces and label them in order of decreasing potential for: a) one positive charge; b)

one negative charge; c) two identical positive point charges at a distance d from each other; and d) a negatively charged infinitely large metal plate.

29. Show a charge arrangement and a point in space where the absolute potential produced by the charges is

zero but the E!

field is not zero. 30. (a) Explain what happens if you place a conductor in an external electric field. How can you test your

explanation? (b) What happens if you cut the conductor in half while keeping it in the external electric field? How can you test your answer?

31. (a) Explain what happens if you place a dielectric in an external electric field? (b) What happens if you cut

a dielectric in half while keeping it in the external electric field? How can you test your answer? 32. How can you prove that there is no electric field inside conductors? 33. Explain why the charge on an electrical conducting material is located on its surface. Problems In some of these problems you may need to know that the mass of an electron is 9.11 x 10-31 kg and the mass of a proton is 1.67 x 10-27 kg. 15.1 and 15.2 Mechanisms of electrostatic interactions and the physical quantity E

! field

1. (a) Construct a graph of the magnitude of the E!

field-versus-position for the electric field created by a point like object with charge Q. . (b) Using the same set of axes draw a graph for the field produced by an object of charge 2Q. . (c) Using the same set of axes draw a graph for the field produced by an object of charge 2Q+ .

2. A uranium nucleus has 92 protons. (a) Determine the magnitude of the E

! field at a distance of

-120.58 10 m, from the nucleus (about the radius of the innermost electron orbiting the nucleus). (b)

What is the magnitude of the force exerted on an electron due to this E!

field? (c) What assumption did you make? If the assumption is not valid, will the magnitude of the force increase or decrease?

3. Two objects with charges -9+4.0 10 C, and -9+3.0 10 C, are 50 cm from each other. Find a

location where the E!

field due to these two charged objects is zero. Draw electric field lines for the situation.

4. The electron and the proton in a hydrogen atom are about 10-10 m from each other. What quantities related

to the electric field can you determine using this information?


5. A -9+4.0 10 C, charged object is 4 cm along a horizontal line toward the right of a -93.0 10 C+ ,

charged object. (a) Determine the E!

field at a point 3.0-cm directly above the negative charge. (b) What else can you find using the information provided?

6. Use superposition principle to draw electric field lines for the following arrangements of charged objects.

Consider all objects to be point-like and decide what distance you want between them: (a) q. and q. ; (b) q. and 3q. , (c) q. and q+ ; (d) q. and 3q+ ; (d) q. , q. and 3q+ .

7. Use the information in Problem 6 to show how you can determine the value of some electric physical

quantity in each of the situations. What additional information will you need? 8. Electric field lines for a field created by an arrangement of charged objects are shown in Fig. P15.8. (a)

Where are these charged objects located and what are the signs of there electric charge? (b) What else can you determine using the information? Give two examples.

Figure P15.8

15.3 Skills for E!

Field Problems 9. A 3.0-g aluminum foil bit with a charge of -9+4.0 10 C, is suspended on a string in a uniform

horizontal electric field. The string makes an angle of 030 with the vertical. What information about the electric field can you determine for this situation?

10. (a) If the string in the previous problem is cut, how long will it take the bit to fall to the floor 1.5 m

below the level of the bit? (b) How far will the bit travel in the horizontal direction while falling? Indicate all the assumptions that you made.

11. Using the Earth’s electric field for flight? The Earth has an electric charge on its surface that produces a

150 N/C E!

field that near its surface points down toward the center of the Earth. Estimate the electric charge that a person would need so that the electric force that this electric field exerts on the person will support the person in the air above the Earth. Indicate all of your assumptions. Is this a reasonable idea? Explain.

12. Two objects each with charge q. are separated by a distance d . (a) Determine the value of the E

! field

at a point that is a distance d from both charges; (b) What else can you determine using the information provided in the problem statement? Give two examples.

13. An electron is projected toward the right with a horizontal velocity of magnitude 0v into a constant

downward E!

field between two parallel plates. Derive an equation for its trajectory while in the field. 14. An electron moving with a speed 0v enters a region where an E

! field points in the same direction as the

electron’s velocity. What will happen to the electron in this field? Answer the question qualitatively and quantitatively (using symbols).


15. A point-like charged object with a charge +q is placed in an external horizontal uniform electric field E!

o that points to the right. Determine an expression for the net E

! field at a distance d from the charge: (a) to

the right of it and along a horizontal line going through the charge q. ; and (b) along a vertical line going through the charge q. and above it.

16. A 1.0-g aluminum foil ball with a charge of -91.0 10 C, hangs freely from a 1.0-m long thread. What

happens to the ball when a horizontal 5000 N/C electric field is turned on? Answer the question as fully as possible.

17. A 0.50-g oil droplet with charge -95.0 10 C. , is in a vertical E

! field. In what direction should the

E!

field point and what magnitude should it have so that the droplet moves at constant speed? Indicate all of the assumptions that you made. If the assumptions are not valid, how would the answer change?

18. An electron is ejected into a horizontal uniform E

! field at a horizontal velocity of 1000 m/s . Describe

everything you can about the motion of the electron. What reasonable assumptions can you make to simplify your description?

19. Two small aluminum foil bits are 50 cm from each other and exert 0.10-N electric forces on each other.

Estimate the magnitude of the electric field at a point half way between the bits. Indicate all assumptions that you made and describe how the answer will change (increase or decrease) if the assumptions are not valid.

20. Equation Jeopardy Problem 1 The equations below describe one or more physical processes. Solve the

equations for the unknowns and write a problem statement for a problem for which the equations are a satisfactory solution.

0

-6 0

-(0.10 kg)(9.8 N/kg) + sin 53 = 0+(-1.0 10 C) + cos 53 = 0x

TE T,

21. Equation Jeopardy Problem 2 The equations below describe one or more physical processes. Solve the

equations for the unknowns and write a problem statement for a problem for which the equations are a satisfactory solution.

-5 4(+1.0 10 C)(-4.0 10 N/C) = (0.20 kg) 0 = (8.0 m/s) +

x

x

aa t

, ,

15.4 and 5 Physical quantity: V field (electric potential) and equipotential surfaces 22. During a lightning flash, -15 C of charge moves through a potential difference of 78.0 10 V, .

Determine the change in electrical potential energy of the field-charge system. 23. (a) Sketch a V -versus-position graph for the electric potential created by a point like object with charge

Q. . (b) Using the same set of axes, draw a graph for an object with charge 2Q. . (c) Using the same set of axes draw a graph for an object with charge –2Q .

24. Two objects each with charge q. are separated by a horizontal distance d . Determine the value of the V

field at a point that is located at a distance d from each charge. 25. Two objects of the same magnitude charge q and different signs are separated by a distance d . Determine

the value of the V field at a point that is located at a distance d from each charge.


26. Four objects with the same negative charge q+ are placed at the corners of a square of side d . Determine

the values of the E!

field and V field in the center of the square. 27. Spark jumps to nose An electric spark jumps from a person's finger to your nose and through your body.

While passing through the air, the spark travels across a potential difference of 42.0 10 V, and releases 73.0 10 J+, of electrical potential energy. How much charge in coulombs and how many electrons flow?

28. Draw equipotential surfaces for the following charge distributions. For each case, make the potential

difference between adjacent surfaces (lines when drawn on a paper) the same magnitude and label them in the order of decreasing potential: a) one object with positive charge q. ; b) one object with negative charge q+ ; c) two identical positively charged objects q. at a distance d from each other; and d) a negatively charged infinitely large metal plate.

29. Two -6–3.0 10 C, charged point-like objects are separated by 0.20 m . Determine the absolute

potential (assuming zero volts at infinity) at a point (a) halfway between the objects and (b) 0.20 m to the side of one of the objects (and 0.40 m from the other) along a line joining them.

30. For a system of two equal magnitude oppositely charged objects, determine the location of a position

where: (a) the magnitude of the E!

-field is zero; and (b) the V -field (potential) is zero. 15.6 and 15.7 Skills for V Field (Electric Potential) Problems and Relating the E

! field and the Electric

Potential V 31. The potential difference across the tungsten-heating element of a toaster is 120 V . Electrons flowing

along the wire have no change in kinetic energy. All of the electrical potential energy loss of the battery and moving electrons system is converted to thermal energy due collisions of the moving electrons with the core ions in the tungsten. Determine the number of electrons that must flow across the 120 V potential difference each second to produce 500 J of thermal energy each second.

32. A 0.10-kg cart with a charge of -5+6.0 10 C, rests on a 037 incline. Through what potential

difference must the cart pass so that it travels 2.0 m up the incline before it stops? The cart starts at rest. Assume that the gravitational constant is 10 N/kg .

33. You have a Van De Graff generator charged to -8+1.0 10 C, and a positively charged aluminum ball

with the charge of -9+1.0 10 C, . What problem involving electric potential can you solve using this information?

34. The potential difference from the cathode (negative electrode) to the screen of an old television set is

+22,000 V . An electron leaves the cathode with an initial speed of zero. What can you determine about the motion of the electron in the TV set using this information?

35. A 10,000-kg shuttle bus carries a +15-C charge on a sphere on its top. Through what potential

difference must the bus travel to acquire a speed of 10 m/s , assuming no friction force exerted on the bus? 36. A charged conductor is at a potential of +300 V . Determine the minimum speed of an electron just

outside the conductor’s surface that it needs to be able to fly to infinity where the electric potential is zero.


37. The electric field on the earth points outward and has a magnitude of about 100 V/m . Determine the potential difference between the ground and your bedroom window 10 m above the ground. Describe the assumptions that you made in your calculations. How can you make sure that these assumptions are valid?

38. Electric field in body cell The electric potential difference across the membrane of a body cell is

+0.70 V (higher on the outside than on the inside). The cell wall is -98.0 10 m, thick. Determine the magnitude and direction of the electric field through the cell wall. Describe any assumptions you made.

39. Energy used to charge nerve cells A nerve cell is shaped like a cylinder. The membrane wall of the

cylinder has a +0.70 V potential difference from the inside to the outside of the wall. To help maintain this potential difference, sodium ions are pumped from inside the cell to the outside. For a typical cell, 109 ions are pumped each second. (a) Determine the change in chemical energy each second required to produce this increase in electrical potential energy. (b) If there are roughly 117 10, of these cells in the body, how much chemical energy is used in pumping sodium ions each second. (c) Estimate the fraction of a person's metabolic rate used to pump these ions.

40. Protons are accelerated from the end of an ion gun used to propel a rocket ship. (a) Determine their speed if

they start at rest and pass through a potential difference of -50,000 V . (b) Determine the momentum that 3.0 kg of these protons acquire. (c) Determine the final speed of a 4000-kg rocket ship after emitting the protons. Assume that the ship starts at rest while floating in space and that all the protons are emitted in one burst from the resting ship. (d) Will this method of propulsion cause any problems for the rocket or for its occupants? Explain.

41. You hold two positively charged marbles with charges 1q and 2q at a distance d apart on a smooth table.

What happens to the marbles if you let them go? Draw an energy bar chart for the process (specify the system). Find the speed of the marbles after the distance between them increases to 2d . List all the assumptions that you made.

42. Two positively charged marbles with the charges 1q and 2q are connected by a spring with the spring

constant k and placed on a smooth table. If the length of the spring before being connected to the marbles is 0l , then what is the spring length after it is connected to the marbles?

43. Equation Jeopardy Problem 3 The equation below describes one or more physical processes. Solve the

equation for the unknown and write a problem statement for a problem for which the equation is a satisfactory solution.

2 -40 = (1/2)(100 kg)(6.0 m/s) + (2.0 10 C) V, 2 . 44. Equation Jeopardy Problem 4 The equation below describes one or more physical situations. Solve the

equation for the unknown and write a problem statement for a problem for which the equation is a satisfactory solution.

9 2 2 -5

9 2 2 -5

(9.0 10 Nm /C )(+2.0 10 C)/(2000 m)+ (9.0 10 Nm /C )( – 2.0 10 C)/(1000 m) = V

, ,

, , 2

45. Can electrically charged objects be arranged so that the E

! field that they produce and the absolute

potential V are both zero at the same point? Show such an arrangement of charges and a point or points that satisfy this condition, that is, if you find such a point.


46. Show a charge arrangement and a point in space where the absolute potential produced by the charges is zero but the E

! field is not zero.

15.8 and 15.9 Conductors and dielectrics in an electric field 47. A metal sphere has no charge on it. A positively charged object is brought near, but does not touch the

sphere. Show that this object can exert a force on the sphere even though the sphere has no net charge. How can you test your answer experimentally?

48. Suppose a Van de Graf generator of radius 0.10 m has a charge of about -6–1.0 10 C, on it. The Van

de Graf generator is than turned off and grounded. How many excess electrons remain on its dome? 49. A metal ball of radius 1R has a charge Q . Later it is connected to a metal ball of radius 2R . What is the

fraction of the charge Q that remains on the first ball? 50. Draw an E

! field-versus-position graph for the field caused by a positively charged metal sphere of radius

R . 51. Draw a V field-versus-position graph for the field caused by a positively charged hollow metal sphere of

radius R . 52. A small charged object A (charge q. ) is placed inside a metal sphere of radius R . Another charged

object B is a distance r form the first one ( r R< ). (a) Determine the magnitude and the direction of the electric force exerted on A. (b) Determine the magnitude and the direction of the electric force exerted on B. (c) Compare the forces exerted on each object and explain whether the result violates Newton’s third law.

53. (a) Draw a picture with a microscopic representation of what happens to the charge distribution in a

conductor placed in an external electric field. For simplicity make the conductor of some regular shape. How can you test your explanation? (b) What happens if you cut the conductor in half while keeping it in the external field? How can you test your answer?

54. (a) Draw a picture with a microscopic representation of what happens if you place a dielectric material in

an external electric field. (b) What happens if you cut the dielectric in half while keeping it in the external field? How can you test your answer?

55. A metal box is placed in a uniform electric field. It is oriented in an arbitrary direction with the respect to

the field lines of that uniform field. Show the direction of the electric field lines outside and near the box and inside the box.

56. An infinitely long plate of conducting material is charged with a charge Q . (a) Draw electric field lines for

the plate. (b) A small metal sphere with the charge of q is placed at a distance d from the plate. Show the new distribution of the field lines. (c) Use the distribution of the lines to estimate the force that the plate exerts on the small sphere. (Hint: think of what arrangement of two charged spheres could produce the same line distribution as the plate and the sphere.)

57. Describe an experiment that you can perform to test if there is no electric field inside a hollow conductor

when it is in an external electric field. 58. Describe an experiment that you can perform to test if the electric field is not zero inside a dielectric

material in an external electric field.


59. It is possible to charge an electroscope without touching it with a charged object. To do this, you need to bring a charged object to the top of the electroscope and while holding it there, touch the top with your other hand and only after that you should remove your hand and then the charged object. You will see the electroscope registering some charge. (a) Where did the charge come from? Explain and draw a microscopic picture of the process. (b) If the charged object you held near by had a negative charge, what is the sign of the charge on the electroscope now?

60. A positively charged metal ball A is placed near metal ball B. Measurements demonstrate that the force that

they exert on each other is zero. Is ball B charged? Explain. 61. Positively charged metal ball A is placed near metal ball B. Ball B has a very small positive charge.

Explain why the balls could attract each other. Draw sketches of the balls and their charge distribution to support your answer.

62. Two small metal spheres A and B have different electric potentials (A is at higher potential). Describe in

words and mathematically what happens if you connect them with a wire. What assumptions did you make? How will the answer change if the assumptions are not valid?

63. Two metal balls of the same radius are placed close to each other. (a) Will they exert the same magnitude

force on each other when they have like charges as when they have opposite charges? Explain. (b) What assumptions did you make? If you made different assumptions, would your answer be different? (c) How will you set up an experiment to test your answer?

64. A wire connects a charged metal ball to an electroscope. How will the electroscope’s reading change if you

move the ball closer to a grounded metal pipe? Explain. 65. An electric dipole such as a water molecule is in a constant E

! field. (a) Will the force exerted by the field

cause the dipole to have a linear acceleration along a line in the direction of E!

? Explain. (b) Will the field exert a torque on the dipole? Explain.

66. Why does an electroscope discharge faster by itself if in humid weather than in dry weather? 67. Imagine that your left hand touches a metal rack at 0 V , and your right hand accidentally touches the

100-V power source in a piece of electronic equipment. Indicate the direction in which sodium ions in the tissue of your right arm flow. In which direction do chlorine ions flow? Justify your answer. (Do not try this experiment.)

68. Electric field of a fish An African fish called the Gymnarchusniloticus has a

charge -7+1.0 10 Cq " , at its head and an equal magnitude negative charge q+ at its tail (see Fig. P15.68). Determine the magnitude and direction of the

electrical field at position A, and the force on a hydroxide ion (charge e+ ) at that point. Remember that the fish and ion are in water.

69. Body cell membrane electric field (a) Determine the average magnitude of the E

! field across a body cell

membrane. A 0.075-V potential difference exists from one side to the other, and the membrane is -97.5 10, m thick. (b) Determine the magnitude of the electrical force on a sodium ion (charge e. ) in the

membrane. Assume that the dielectric constant is 1.0 (it is really somewhat larger). 70. Earth’s electric field Earth has an electric charge of approximately 55.7 10 C, distributed relatively

uniformly on its surface. Determine as many quantities as possible describing the electrical properties of the space around Earth using this and any additional information that you need.


71. Two small metal spheres are attached to the ends of a thin, 1.6-m -long glass rod that fractures if compressed by a force greater than 640 N . If the spheres and rod are immersed in benzene, what is the maximum charge that can be transferred from one sphere to the other before the rod fractures?

72. Geological exploration Determine the electric field at position A in Fig.

P15.72 due to the dipole charges produced by a geologist's electrodes. The dielectric constant of the soil is 8.0 and the dipole charge –34.010 Cq " . Determine the force exerted by this field on a sodium ion (charge e) at point A.

15.9 Capacitors 73. A parallel-plate capacitor is an electronic device that consists of two large parallel plates separated by a

small distance. (a) Determine the average E!

field between the plates if a 120-V potential difference exists from one plate to the other and if their separation is 0.50 cm. (b) A spark will jump if the magnitude of the E

! field exceeds 3.0 x 106 V/m, assuming air separates the plates. What is the closest the plates can be

placed to each other without sparking? 74. A capacitor of capacitance C with a vacuum between the plates is connected to a potential difference .V2

(a) Write an expression for the charge on each of the plates. (b) For the total energy stored by the capacitor. (c) The capacitor is now filled with a dielectric material with a dielectric constant 9 (being connected to the same potential difference). What is the new charge on the plates? (d) What is the new energy stored by the capacitor? (e) Where did the energy change come from?

75. A capacitor of capacitance C with a vacuum between the plates is connected to a potential difference source

.V2 (a) Write an expression for the charge on each of the plates. (b) For the total energy stored by the capacitor. (c) The capacitor is now disconnected from the potential difference source and filled with a dielectric with a dielectric constant 9 . What is the new charge on the plates? (d) What is the new energy stored by the capacitor? (e) Where did the energy change come from or where did the energy go?

76. Use the superposition principle to show that if a capacitor has infinitely large plates, the electric field is

concentrated between the plates and not outside the plates. 77. How does the capacitance of a parallel-plate capacitor change if you double the magnitude of the charge on

its plates? If you triple the potential difference across the plates? What assumptions did you make? 78. Axon capacitance The long thin cylindrical axon of a nerve carries nerve impulses. The axon can be as

long as 1 m. (a) Estimate the capacitance of a 1.0-m long axon of radius of 4.0 x 10-6 m with a membrane thickness of 8.0 x 10-9 m. The dielectric constant of the membrane material is about 6.0. (b) Determine the magnitude of the charge on the inside (negative) and outside (positive) of the membrane wall if there is a 0.070-V potential difference across the wall. (c) Determine the energy stored in this axon capacitor when charged.

79. Cow capacitance A metal sphere of radius R has an electric charge q. on it. Determine the electric

potential V on the sphere’s surface. Use the definition of capacitance to show that the capacitance of this

isolated sphere is Rk

where k is the constant used in Coulomb’s law. Estimate the capacitance of a

“spherical cow”, a favorite subject of physicists. 80. You have a capacitor of the capacitance C with plate separation d and filled with dielectric 9 . Design

three different problems that you can solve using this information. Then solve the problems to make sure you have enough information.


81. Capacitance of red blood cell Assume that a red blood cell is spherical with a radius of 4 x 10-6 m and

with wall thickness –89 10 m, . The dielectric constant of the membrane is about 5. Assuming the cell is a parallel plate capacitor, estimate the capacitance of the cell and calculate the positive charge on the outside and the equal magnitude negative charge inside when the potential difference across the membrane is 0.080 V .

82. Defibrillator Ventricular fibrillation is the cause of some heart attacks. The heart muscles contract

randomly preventing the coordinated pumping of blood. A defibrillator can often restore normal blood pumping by discharging the charge on a capacitor through the heart. Paddles held against the patient’s chest and a -66 10 F, capacitor charged to high voltage is discharged in several milliseconds. If the capacitor energy is 250 J , what potential difference is used to charge the capacitor?

15.10 Putting it all together 83. Another spark If the magnitude of the E

! field between your finger and some other object exceeds

63.0 10 V/m, , a spark will jump. If you shuffle across a rug, a spark will jump when your finger is 0.20 cm from another person's nose. Estimate the potential difference between your finger and the other person's nose just before the spark jumps.

84. The dielectric strength of air is 63 10 V/m, . As you walk across a synthetic rug, your body accumulates

electric charge, causing a potential difference of 6000 V between it and a doorknob. What can you can determine using this information?

85. Charged cloud causes electric field on Earth The electric charge on clouds is a complex subject.

Consider the simplified model shown in Fig. 15.47a. A positive charge is near the top of the cloud and a negative charge near the bottom. Determine the direction of the electric field on the Earth at point P below the cloud and explain so a classmate can understand why there is positive charge on the ground directly below the cloud.

86. Heart’s dipole charge The heart has a dipole charge distribution with a charge of + 1.0 x 10-7 C 6.0 cm

above a charge of –1.0 x 10-7 C. Determine the electric field (magnitude and direction) caused by the heart's dipole at a distance of 8.0 cm directly above the heart's positive charge. All charges are located in body tissue of dielectric constant 7.0. What is the force exerted on a sodium ion (charge +1.6 x 10-19 C) at that point?

87. More heart dipole charge Repeat the previous problem only now for a point in the tissue that is 4.0 cm

horizontally to the left side of the middle point between the two heart charges. General Problems 88. A spec of dust of the mass 0.1 µg floats between two oppositely charged horizontal metal plates. The

potential difference between the plates is 200 V. The spec is not falling at first, but when a beam of ultraviolet light hits it, it starts accelerating downward. However, by increasing the potential difference to 250 V, the downward acceleration of the spec becomes zero. Explain the phenomenon qualitatively and quantitatively using symbols.

89. A 0.050 kg cart has a dome that is charged to 1.0 x 10–5 C. The cart is at the top of a 2.0-m long inclined

plane that makes a 300 angle with the horizontal. If the system is placed in a vertical uniform 4000 N/C electric field, how long will it take the cart to reach the bottom of the plane? What assumptions did you make?

90. Can shark detect axon electric field? A nerve signal is transmitted along the long thin axon of a neuron in

a small fish. The transmission occurs as potassium ions K+ transfer like tipping dominos across the axon membrane from inside to outside. Each short section of axon gets an excess of about 6 x 108 potassium


ions/mm. Determine the electric field 4.0 cm from the axon produced by the excess potassium ions on the outside of the axon and an equal number of larger in size negative ions on the inside of a 1-mm length of axon. The ions are separated by the 8 x 10-9 m thick axon membrane. Will a shark that is able to detect fields as small as 10-6 N/C be able to detect that axon field? Explain.

91. Two conductors are charged positively. One has a potential of +100V and the other’s is +50 V. Will the

electric charge move from the first one to the second one if you make them touch each other? Consider three cases: a) both conductors are small balls; b) the first conductor is a large hollow sphere with an opening and the second one is a small metal ball; c) the second conductor is a hollow sphere with an opening and the first one is a small metal ball. Consider all possibilities and justify your answers with qualitative and quantitative reasoning.

92. Lightning warms water A lightning flash occurs when –40 C of charge moves from a cloud to Earth

through a potential difference of 4.0 x 108 V. How much water can boil due to energy released during the process? Make necessary assumptions for the estimate.

93. In a hot water heater, water warms when electrical potential energy is converted into thermal energy. (a)

Determine the energy needed to warm 180 kg of water by 10 C0. (b) If -10.0 C of electric charge passes through a +120-V potential difference in the heating coils each second, determine the time needed to warm the water by 10 C0.

94. Electrophoresis Electrophoresis is used to separate biological molecules of different dimensions and

electric charge (the molecules can have different charge depending on the Ph of the solution). A particular molecule of radius R with charge q is in a viscous solution, which has an electric field across it. The field exerts an electric force on the molecule and the viscous solution exerts an opposing drag force

dragF DRv" , where D is a constant drag coefficient that depends on the shape and other features of the molecule and the solution, and v is the molecules speed in the solution. When the molecule gets up to speed, the electric force exerted on it by the field is equal in magnitude and opposite in direction to the drag

force. Show that during time interval t2 , the molecule will travel a distance qE tx

DR2

2 " . Describe any

assumptions you made. 95. Energy stored in axon electric field An axon has a surface area of -6 25 10 m, and the membrane is

-98 10 m, thick. The dielectric constant of the membrane is 6. (a) Determine the capacitance of the axon considered as a parallel plate capacitor. (b) If the potential difference across the membrane wall is 0.080 V , determine the magnitude of the charge on each wall. (c) Determine the energy needed to charge that axon capacitor. (d) Determine the magnitude of the electric field across the membrane due to the opposite sign charges across the membrane walls. (e) Calculate the energy density of that electric field. (f) Multiply the volume of space occupied by that electric field by the volume of the membrane to get the total energy stored in the electric field. How do the answers to c and f compare?

96. For a system of two oppositely charged objects of arbitrary magnitude charge, derive a general expression

to determine the location of a position where: (a) the E!

-field is zero; (b) the V -field potential is zero. 97. An infinitely long plate of conducting material is charged with a charge Q . A small metal sphere with the

charge of q is placed at a distance d from the plate. Estimate the force that the plate exerts on the sphere. Reading Passages 1. Electric Discharge by Eels In several sea animals, such as the Nile catfish and the South American electric eel, their electric organs produce 600-V potential difference pulses to ward off predators as well as to stun


prey.* Figure 15.48 illustrates the key component that produces this electric shock—an electrocyte. When the electrocyte is inactive as in (a), its interior has an excess of negatively charged potassium ions (K–). The exterior has an excess of positively charged sodium ions (Na+). There is a –90 mV electric potential difference from outside the cell wall to inside the electrocyte cell wall, but zero potential from one exterior side to the other exterior side. How then does an eel produce the 600-V potential difference to stun an intruder?

Figure 15.48 Electric eel shocking system

First, many electrocytes are placed one after the other in columns in the eel’s long trunk and tail (Fig.

15.48c). Each electrocyte has several types of ion channels. When activated by a special molecule, a channel allows a certain type of ion to pass through the channel from outside the cell to the inside, or vice versa. The activation starts with a nerve impulse originating in the eel’s brain that travels down its spinal column and causes the emission of the chemical acetocholine at the flat sides of the electrocytes. The acetocholine opens a sodium channel that causes external sodium ions to pass from outside the cell wall to the inside, but only on the left flat side of the electrocyte. This causes the electric potential across that cell wall to change from –90 mV to +50 mV . The electric potential from the left external side to the right external side of an electrocyte is now about 100 mV (Fig. 15.48c). With about 6000 electrocytes in series, the release of acetocholine causes an electric impulse of over 600 V (6000 electrocytes in series times 0.10 V per electrocyte). The discharge lasts about 2-3 ms .

98. Suppose you have a 1.0-F capacitor (very large capacitance) with a 0.10-V potential difference across

the capacitor. What is the magnitude of the electric charge on each plate of the capacitor? (a) 0.05 C (b) 0.10 C (c) 0.20 C (d) 1.0 C (e) 10 C

99. Suppose you place two of these 1.0-F capacitors with the charge calculated

in the previous question as shown in Fig. P15.99a. What is the net potential difference across the two capacitors (from one dot to the other)?

(a) 0.05 V (b) 0.10 V (c) 0.20 V (d) None of these. (e) No clue.

100. If both capacitors are discharged simultaneously, how much electric charge goes through the wire shown in

Fig. P15.99a? (a) 0.05 C (b) 0.10 C (c) 0.20 C (d) 1.0 C (e) 10 C

101. Suppose you place the two capacitors with the 0.10-V potential difference across

each capacitor as shown in Fig. P15.99b. What is the net potential difference across the two capacitors?

(a) 0.05 V (b) 0.10 V (c) 0.20 V (d) None of these. (e) No clue.

* J. Bastian, “Electrosensory Organism,” pp. 30-37, Physics Today 47 (2), 1994.


102. Look at the electrocyte shown in Fig. 15.48c. What causes the 0.10 V potential difference from the outer left to the outer right side of the cell? * (a) The membrane is thicker on the left than on the right. (b) The ion distribution across the left wall is different than across the right wall. (c) The left and right membranes have different capacitances. (d) b and c. (e) a, b, and c.

103. Suppose that one cell of the electrocyte is regarded as a small capacitor with a 0.10 V potential

difference across it. How should we arrange ten cells to get a 1.0 V potential difference across them? (a) In series. (b) In parallel. (c) Randomly so that they do not cancel each other. (d) No clue.

2. Electrostatic precipitator (ESP), or electrostatic air cleaner These are devices that remove particles from a flowing gas, such as air. ESPs are used to remove particle emissions from smoke moving up through smokestacks in coal and oil fired electricity-generating plants, and pollutants from the boilers in pulp mills and in oil refineries. You can buy portable versions or whole-house models that connect to the cold-air return on the furnace. They remove about 95% of dirt and 85% of microscopic particles.

The most basic precipitators contain a negatively charged metal grid (thin wires) followed by a stack of large flat vertically oriented metal plates, with the plates typically spaced about 1 cm apart (depicted in Fig. 15.49). The air flows across the charged grid of wires, and then passes between the stack of plates. A large negative voltage (tens of thousands of volts) is applied between the wires and the plates. Electric discharge (sparks) ionizes the air around the thin wires. Negative ions join the gas particles that flow upward between the plates. The ionized particles are attracted to and stick to the oppositely charged plates, thus removing them from the moving gas.

Figure 15.49 Electrostatic precipitator

104. It takes about a 63 10 V/m, electric field to ionize dry air molecules. If the potential difference between the thin wires and metal plates is 60,000 V , what minimum distance must the wires be from the plates in order to cause dry air to ionize? (a) 1.0 cm (b) 2.0 cm (c) 4.0 cm (d) 50 cm (e) 50 m

105. The smoke particles are attracted to the closely spaced plates because:

(a) The particles are negatively charged. (b) The particles are positively charged.

* The right side of the electrocyte now has an excess of positive charge and the left side an excess of negative charge. It is this unbalanced charge distribution that produces the potential difference—and also an E field between the charges. Chemical processes inside a battery also produce separation of charges (more negative charge on one battery terminal than on the other). This charge separation is responsible for the potential difference across the battery terminals (12 V for a car battery and 1.5 V for a flashlight battery).


(c) The plates are positively charged. (d) The plates are negatively charged. (e) a and c are correct. (f) b and c are correct.

106. Suppose a -62.0 10 kg, dust particle with charge -9–4.0 10 C, is moving straight up a chimney at

speed 6.0 m/s when it enters the +2,000 N/C electric field pointing away from a metal collection plate of an electrostatic precipitator. If the particle is 2.0 cm from the plate at that instant, which answer below is closest to the magnitude of its horizontal acceleration toward the plate? (a) -5 22.0 10 m/s, (b) -4 21.2 10 m/s, (c) -4 24.0 10 m/s, (d) 24.0 m/s (e) 224 m/s

107. Suppose a -62.0 10 kg, dust particle with charge -9–4.0 10 C, is moving straight up a chimney at

speed 6.0 m/s when it enters the +2,000 N/C electric field pointing away from a metal collection plate of an electrostatic precipitator. If the particle is 2.0 cm from the plate at that instant, which answer below is closest to the vertical distance it will move before hitting the plate? (a) 0.01 m (b) 0.02 m (c) 0.04 m (d) 0.06 m (e) 0.1 m

108. Why are the particles collected by the precipitator given a negative electric charge?

(a) They are then attracted to the positively charged plates while moving upward. (b) They are then repelled from the positively charged plates while moving upward. (c) The extra mass slows their upward movement. (d) This cancels the positive charge they have when entering the precipitator. (e) a and c are correct.

109. The particles move up a chimney between metal plates. Why are the plates charged positively?

(a) They attract the positively charged particles while moving upward. (b) They attract the negatively charged particles while moving upward. (c) This prevents the plates from collapsing into each other. (d) This causes the particles collected on the surface to fall into collection bins. (e) This causes the particles to become positively charged.

110. Suppose the average particle moving with air up a chimney has a mass of -64.0 10 kg, and charge

-8–6.0 10 C, . They move with the air at speed 6.0 m/s and on the average 3.0 cm from a vertical plate. The plate is 0.60-m tall. Which answer below is closest to the minimum horizontal electric field at the plate be so that the particles will be collected before moving past the top of the plate? (a) 100 N/C (b) 200 N/C (c) 400 N/C (d) 1000 N/C (e) 4000 N/C

Answers to Review Questions 15.1 According to the action-at-a distance model, two electrically charged objects exert electric forces on each other without any mediator. If one of the objects moves, the other “knows” about it immediately as no time is needed to adjust the force between them. According to electric field model, each electrically charged object produces an electric field that surrounds itself and the field exerts a force on any other charged objects placed near by.

15.2 Draw a picture showing E!

field vectors due to each charge at the point of interest. Use 2sourceqE kr

" for

each source charge to calculate the magnitude of the E!

field due to each of them at the point of interest. Now


you know the direction and the magnitude of the two E!

field vectors. Add them as vectors using head to tail method and find the magnitude of the sum. 15.3 The magnitude of the E

! field of a point-like object decreases proportionally to the square of the distance;

the magnitude of the E!

field of the infinitely large plate does not depend on the distance from the plate. The electric field lines of the point-like object are radial, the lines of the uniformly charged pate are perpendicular to the plate and parallel to each other. 15.4 Place a small positive electrically charged object at the point of interest. Then calculate the total electrical potential energy of a system that includes that object and all other charged objects. Finally, divide the total electric potential energy by the charge placed at that point. The is the electric potential at that point. 15.5 In both cases the work is zero. The circular path is an equipotential surface. 15.6 A positively charged object will accelerate left to right and the negatively charged object right to left. 15.7 If the E

! field at some location is zero, it means that in that vicinity, the electric potential does not change;

it can be a zero or some other constant value. 15.8 In the metal can on top of the electroscope experiment, the free electrons redistribute themselves in the presence of the electric field created by the negatively charged object and create their own E

!induced field

opposite the direction and of the same magnitude as the E!

field of the charged object. Thus, inside the can the net electric field is zero and the electrons in the electroscope needle do not move to the bottom of the needle. 15.9 Inside a conducting material placed in an external electric field the net electric field is zero; inside a dielectric material the field is weaker but not entirely zero. The difference can be explained microscopically. In the conductors there are freely moving charged particles that can redistribute to cancel the external field. Inside a dielectric, the molecules change their shape and orientation but do not move translationally; thus the field is diminished but not cancelled. 15.10 Place a potential difference (for example, from a battery) across the plates. 15.11 The heart’s dipole charge causes positive and negative ions in the tissue to travel in opposite directions. The magnitude and orientation of that heart dipole thus affects how ions move and where excess positive negative ions reach the electrodes. The electrode signal reflects the heart’s changing dipole charge and hence the number of muscle cells contracting at different times in the heart beat cycle.

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Chapter 15: Electric Field: Force and Energy Approaches