Problems

Section 18.1 The Origin of Electricity Section 18.2 Charged Objects and the Electric Force Section 18.3 Conductors and Insulators Section 18.4 Charging by Contact and by Induction

1. Iron atoms have been detected in the sun's outer atmosphere, some with many of their electrons stripped away.

What is the net electric charge (in coulombs) of an iron atom with 26 protons and 7 electrons? Be sure to include the

algebraic sign ( or ) in your answer.

Answer:

REASONING The charge of a single proton is , and the charge of a single electron is , where

. The net charge of the ionized atom is the sum of the charges of its constituent protons and

electrons.

SOLUTION The ionized atom has 26 protons and 7 electrons, so its net electric charge is

2. An object has a charge of . How many electrons must be removed so that the charge becomes ?

3. Four identical metallic objects carry the following charges: , , , and . The objects are

brought simultaneously into contact, so that each touches the others. Then they are separated.

(a) What is the final charge on each object?

Answer:

(b) How many electrons (or protons) make up the final charge on each object?

Answer:

4.

Four identical metal spheres have charges of , , , and

.

(a) Two of the spheres are brought together so they touch, and then they are separated. Which spheres are they,

if the final charge on each one is ?

(b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each

of the three is ?

(c) The final charge on each of the three separated spheres in part (b) is . How many electrons would

have to be added to one of these spheres to make it electrically neutral?

5. Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of . Sphere B carries a

charge of . Sphere C carries no net charge. Spheres A and B are touched together and then separated. Sphere C is

then touched to sphere A and separated from it. Last, sphere C is touched to sphere B and separated from it.

(a) How much charge ends up on sphere C? What is the total charge on the three spheres

Answer:

REASONING Identical conducting spheres equalize their charge upon touching. When spheres A

and B touch, an amount of charge , flows from A and instantaneously neutralizes the charge on B

leaving B momentarily neutral. Then, the remaining amount of charge, equal to , is equally split

between A and B, leaving A and B each with equal amounts of charge . Sphere C is initially neutral, so

when A and C touch, the on A splits equally to give on A and on C. When B and C touch, the

on B and the on C combine to give a total charge of , which is then equally divided between

the spheres B and C; thus, B and C are each left with an amount of charge .

SOLUTION Taking note of the initial values given in the problem statement, and summarizing the final

results determined in the REASONING above, we conclude the following:

(a) Sphere C ends up with an amount of charge equal to .

(b) The charges on the three spheres before they were touched, are, according to the problem statement,

on sphere A, on sphere B, and zero charge on sphere C. Thus, the total charge on the spheres

is .

(c) The charges on the spheres after they are touched are on sphere A, on sphere B, and

on sphere C. Thus, the total charge on the spheres is .

(b) before they are allowed to touch each other and

Answer:

REASONING Identical conducting spheres equalize their charge upon touching. When spheres A

and B touch, an amount of charge , flows from A and instantaneously neutralizes the charge on B

leaving B momentarily neutral. Then, the remaining amount of charge, equal to , is equally split

between A and B, leaving A and B each with equal amounts of charge . Sphere C is initially neutral, so

when A and C touch, the on A splits equally to give on A and on C. When B and C touch, the

on B and the on C combine to give a total charge of , which is then equally divided between

the spheres B and C; thus, B and C are each left with an amount of charge .

SOLUTION Taking note of the initial values given in the problem statement, and summarizing the final

results determined in the REASONING above, we conclude the following:

(a) Sphere C ends up with an amount of charge equal to .

(b) The charges on the three spheres before they were touched, are, according to the problem statement,

on sphere A, on sphere B, and zero charge on sphere C. Thus, the total charge on the spheres

is .

(c) The charges on the spheres after they are touched are on sphere A, on sphere B, and

on sphere C. Thus, the total charge on the spheres is .

(c) after they have touched?

Answer:

REASONING Identical conducting spheres equalize their charge upon touching. When spheres A

and B touch, an amount of charge , flows from A and instantaneously neutralizes the charge on B

leaving B momentarily neutral. Then, the remaining amount of charge, equal to , is equally split

between A and B, leaving A and B each with equal amounts of charge . Sphere C is initially neutral, so

when A and C touch, the on A splits equally to give on A and on C. When B and C touch, the

on B and the on C combine to give a total charge of , which is then equally divided between

the spheres B and C; thus, B and C are each left with an amount of charge .

SOLUTION Taking note of the initial values given in the problem statement, and summarizing the final

results determined in the REASONING above, we conclude the following:

(a) Sphere C ends up with an amount of charge equal to .

(b) The charges on the three spheres before they were touched, are, according to the problem statement,

on sphere A, on sphere B, and zero charge on sphere C. Thus, the total charge on the spheres

is .

(c) The charges on the spheres after they are touched are on sphere A, on sphere B, and

on sphere C. Thus, the total charge on the spheres is .

REASONING Identical conducting spheres equalize their charge upon touching. When spheres A and B

touch, an amount of charge , flows from A and instantaneously neutralizes the charge on B leaving B

momentarily neutral. Then, the remaining amount of charge, equal to , is equally split between A and B, leaving

A and B each with equal amounts of charge . Sphere C is initially neutral, so when A and C touch, the on A

splits equally to give on A and on C. When B and C touch, the on B and the on C combine to give a

total charge of , which is then equally divided between the spheres B and C; thus, B and C are each left with an

amount of charge .

SOLUTION Taking note of the initial values given in the problem statement, and summarizing the final results

determined in the REASONING above, we conclude the following:

(a) Sphere C ends up with an amount of charge equal to .

(b) The charges on the three spheres before they were touched, are, according to the problem statement, on

sphere A, on sphere B, and zero charge on sphere C. Thus, the total charge on the spheres is

.

(c) The charges on the spheres after they are touched are on sphere A, on sphere B, and on

sphere C. Thus, the total charge on the spheres is .

6. A plate carries a charge of , while a rod carries a charge of . How many electrons must be

transferred from the plate to the rod, so that both objects have the same charge?

*7. Water has a mass per mole of 18.0 g/mol, and each water molecule has 10 electrons.

(a)

How many electrons are there in one liter of water?

Answer:

(b) What is the net charge of all these electrons?

Answer:

Section

18.5 Coulomb's Law

8. In a vacuum, two particles have charges of and , where . They are separated by a distance of

0.26 m, and particle 1 experiences an attractive force of 3.4 N. What is (magnitude and sign)?

9. Two spherical objects are separated by a distance that is . The objects are initially electrically

neutral and are very small compared to the distance between them. Each object acquires the same negative charge due

to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of

. How many electrons did it take to produce the charge on one of the objects?

Answer: 8 electrons

REASONING The number of excess electrons on one of the objects is equal to the charge on it divided by

the charge of an electron , or . Since the charge on the object is negative, we can write

, where is the magnitude of the charge. The magnitude of the charge can be found from Coulomb's law

(Equation 18.1), which states that the magnitude of the electrostatic force exerted on each object is given by

, where is the distance between them.

SOLUTION The number of excess electrons on one of the objects is

(1)

To find the magnitude of the charge, we solve Coulomb's law, ,for :

Substituting this result into Equation 1 gives

10. Two tiny conducting spheres are identical and carry charges of and . They are separated by a

distance of 2.50 cm.

(a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?

(b) The spheres are brought into contact and then separated to a distance of 2.50 cm. Determine the magnitude

of the force that each sphere now experiences, and state whether the force is attractive or repulsive.

11. Two very small spheres are initially neutral and separated by a distance of 0.50 m. Suppose that

electrons are removed from one sphere and placed on the other.

(a) What is the magnitude of the electrostatic force that acts on each sphere?

Answer: 0.83 N

REASONING Initially, the two spheres are neutral. Since negative charge is removed from the

sphere which loses electrons, it then carries a net positive charge. Furthermore, the neutral sphere to which

the electrons are added is then negatively charged. Once the charge is transferred, there exists an

electrostatic force on each of the two spheres, the magnitude of which is given by Coulomb's law (Equation

18.1), .

SOLUTION

(a) Since each electron carries a charge of , the amount of negative charge removed

from the first sphere is

Thus, the first sphere carries a charge , while the second sphere carries a charge

. The magnitude of the electrostatic force that acts on each sphere is, therefore,

(b) Since the spheres carry charges of opposite sign, the force is .

(b) Is the force attractive or repulsive? Why?

Answer: attractive

REASONING Initially, the two spheres are neutral. Since negative charge is removed from the

sphere which loses electrons, it then carries a net positive charge. Furthermore, the neutral sphere to which

the electrons are added is then negatively charged. Once the charge is transferred, there exists an

electrostatic force on each of the two spheres, the magnitude of which is given by Coulomb's law (Equation

18.1), .

SOLUTION

(a) Since each electron carries a charge of , the amount of negative charge removed

from the first sphere is

Thus, the first sphere carries a charge , while the second sphere carries a charge

. The magnitude of the electrostatic force that acts on each sphere is, therefore,

(b) Since the spheres carry charges of opposite sign, the force is .

REASONING Initially, the two spheres are neutral. Since negative charge is removed from the sphere which

loses electrons, it then carries a net positive charge. Furthermore, the neutral sphere to which the electrons are added is

then negatively charged. Once the charge is transferred, there exists an electrostatic force on each of the two spheres,

the magnitude of which is given by Coulomb's law (Equation 18.1), .

SOLUTION

(a) Since each electron carries a charge of , the amount of negative charge removed from the

first sphere is

Thus, the first sphere carries a charge , while the second sphere carries a charge

. The magnitude of the electrostatic force that acts on each sphere is, therefore,

(b) Since the spheres carry charges of opposite sign, the force is .

12. Two charges attract each other with a force of 1.5 N. What will be the force if the distance between them is reduced to

one-ninth of its original value?

13. Two point charges are fixed on the y axis: a negative point charge at and a positive

point charge at . A third point charge is fixed at the origin. The net electrostatic

force exerted on the charge q by the other two charges has a magnitude of 27 N and points in the direction. Determine the magnitude of .

Answer:

14.

The drawings show three charges that have the same magnitude but may have different signs. In all cases the

distance d between the charges is the same. The magnitude of the charges is , and the distance between

them is . Determine the magnitude of the net force on charge 2 for each of the three drawings.

15. Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere

is . The gravitational force that each sphere exerts on the other is balanced by the electric force.

(a) What algebraic signs can the charges have?

Answer: the same algebraic signs, both positive or both negative

REASONING AND SOLUTION

(a) Since the gravitational force between the spheres is one of attraction and the electrostatic force must

balance it, the electric force must be one of repulsion. Therefore, the charges must have

.

(b) There are two forces that act on each sphere; they are the gravitational attraction of one sphere for

the other, and the repulsive electric force of one sphere on the other. From the problem statement,

we know that these two forces balance each other, so that . The magnitude of is given

by Newton's law of gravitation (Equation 4.3: ), while the magnitude of is

given by Coulomb's law (Equation 18.1: ). Therefore, we have

since the spheres have the same mass and carry charges of the same magnitude . Solving for ,

we find

(b) Determine the charge magnitude.

Answer:

REASONING AND SOLUTION

(a) Since the gravitational force between the spheres is one of attraction and the electrostatic force must

balance it, the electric force must be one of repulsion. Therefore, the charges must have

.

(b) There are two forces that act on each sphere; they are the gravitational attraction of one sphere for

the other, and the repulsive electric force of one sphere on the other. From the problem statement,

we know that these two forces balance each other, so that . The magnitude of is given

by Newton's law of gravitation (Equation 4.3: ), while the magnitude of is

given by Coulomb's law (Equation 18.1: ). Therefore, we have

since the spheres have the same mass and carry charges of the same magnitude . Solving for ,

we find

REASONING AND SOLUTION

(a) Since the gravitational force between the spheres is one of attraction and the electrostatic force must balance it,

the electric force must be one of repulsion. Therefore, the charges must have

.

(b) There are two forces that act on each sphere; they are the gravitational attraction of one sphere for the other,

and the repulsive electric force of one sphere on the other. From the problem statement, we know that these

two forces balance each other, so that . The magnitude of is given by Newton's law of gravitation

(Equation 4.3: ), while the magnitude of is given by Coulomb's law (Equation 18.1:

). Therefore, we have

since the spheres have the same mass and carry charges of the same magnitude . Solving for , we find

16. A charge is located at the origin, while an identical charge is located on the x axis at . A third

charge of is located on the x axis at such a place that the net electrostatic force on the charge at the origin

doubles, its direction remaining unchanged. Where should the third charge be located?

17. Two particles, with identical positive charges and a separation of , are released from rest.

Immediately after the release, particle 1 has an acceleration whose magnitude is , while

particle 2 has an acceleration whose magnitude is . Particle 1 has a mass of .

Find

(a) the charge on each particle and

Answer:

REASONING Each particle will experience an electrostatic force due to the presence of the other

charge. According to Coulomb's law (Equation 18.1), the magnitude of the force felt by each particle can be

calculated from , where are the respective charges on particles 1 and 2 and is

the distance between them. According to Newton's second law, the magnitude of the force experienced by

each particle is given by , where is the acceleration of the particle and we have assumed that the

electrostatic force is the only force acting.

SOLUTION

(a

) Since the two particles have identical positive charges, , and we have, using the data

for particle 1,

Solving for , we find that

(b

)

Since each particle experiences a force of the same magnitude (From Newton's third law), we can write

, or . Solving this expression for the mass of particle 2, we have

(b) the mass of particle 2.

Answer:

REASONING Each particle will experience an electrostatic force due to the presence of the other

charge. According to Coulomb's law (Equation 18.1), the magnitude of the force felt by each particle can be

calculated from , where are the respective charges on particles 1 and 2 and is

the distance between them. According to Newton's second law, the magnitude of the force experienced by

each particle is given by , where is the acceleration of the particle and we have assumed that the

electrostatic force is the only force acting.

SOLUTION

(a

) Since the two particles have identical positive charges, , and we have, using the data

for particle 1,

Solving for , we find that

(b

)

Since each particle experiences a force of the same magnitude (From Newton's third law), we can write

, or . Solving this expression for the mass of particle 2, we have

REASONING Each particle will experience an electrostatic force due to the presence of the other charge.

According to Coulomb's law (Equation 18.1), the magnitude of the force felt by each particle can be calculated from

, where are the respective charges on particles 1 and 2 and is the distance between

them. According to Newton's second law, the magnitude of the force experienced by each particle is given by

, where is the acceleration of the particle and we have assumed that the electrostatic force is the only force acting.

SOLUTION

(a) Since the two particles have identical positive charges, , and we have, using the data for

particle 1,

Solving for , we find that

(b) Since each particle experiences a force of the same magnitude (From Newton's third law), we can write

, or . Solving this expression for the mass of particle 2, we have

18. A charge of is fixed at the center of a compass. Two additional charges are fixed on the circle of the

compass, which has a radius of 0.100 m. The charges on the circle are at the position due north and

at the position due east. What are the magnitude and direction of the net electrostatic force acting on the

charge at the center? Specify the direction relative to due east.

*19. Multiple-Concept Example 3 provides some pertinent background for this problem. Suppose a single electron orbits

about a nucleus containing two protons , as would be the case for a helium atom from which one of the two

naturally occurring electrons is removed. The radius of the orbit is . Determine the magnitude of the

electron's centripetal acceleration.

Answer:

*20.

The drawing shows an equilateral triangle, each side of which has a length of 2.00 cm. Point charges are fixed

to each corner, as shown. The charge experiences a net force due to the charges and . This net force

points vertically downward and has a magnitude of 405 N. Determine the magnitudes and algebraic signs of the

charges and .

*21. The drawing shows three point charges fixed in place. The charge at the coordinate origin has a value of

; the other two charges have identical magnitudes, but opposite signs: and

.

(a) Determine the net force (magnitude and direction) exerted on by the other two

charges.

Answer:

0.166 N directed along the

(b) If had a mass of 1.50 g and it were free to move, what would be its acceleration?

Answer:

directed along the

*22. An electrically neutral model airplane is flying in a horizontal circle on a 3.0-m guideline, which is nearly parallel

to the ground. The line breaks when the kinetic energy of the plane is 50.0 J. Reconsider the same situation, except

that now there is a point charge of on the plane and a point charge of at the other end of the guideline. In this

case, the line breaks when the kinetic energy of the plane is 51.8 J. Find the magnitude of the charges.

*23. Multiple-Concept Example 3 illustrates several of the concepts that come into play in this problem. A single

electron orbits a lithium nucleus that contains three protons . The radius of the orbit is .

Determine the kinetic energy of the electron.

Answer:

*24.

An unstrained horizontal spring has a length of 0.32 m and a spring constant of 220 N/m. Two small charged

objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of

these charges, the spring stretches by 0.020 m relative to its unstrained length. Determine

(a) the possible algebraic signs and

(b) the magnitude of the charges.

*25. In the rectangle in the drawing, a charge is to be placed at the empty corner to make the net force on the charge

at corner A point along the vertical direction. What charge (magnitude and algebraic sign) must be placed at the empty

corner?

Answer:

REASONING Consider the drawing at the right. It is given that the charges , , and are each positive.

Therefore, the charges and each exert a repulsive force on the charge . As the drawing shows, these forces

have magnitudes (vertically downward) and (horizontally to the left). The unknown charge placed at the

empty corner of the rectangle is , and it exerts a force on that has a magnitude . In order that the net force

acting on point in the vertical direction, the horizontal component of must cancel out the horizontal force

. Therefore, must point as shown in the drawing, which means that it is an attractive force and must be

negative, since is positive.

SOLUTION The basis for our solution is the fact that the horizontal component of must cancel out the horizontal

force . The magnitudes of these forces can be expressed using Coulomb's law , where is the

distance between the charges and . Thus, we have

where we have used the fact that the distance between the charges and is the diagonal of the rectangle, which is

according to the Pythagorean theorem, and the fact that the distance between the charges and is

. The horizontal component of is , which must be equal to , so that we have

The drawing in the REASONING, reveals that . Therefore, we find that

As discussed in the REASONING, the algebraic sign of the charge is .

**26. There are four charges, each with a magnitude of . Two are positive and two are negative. The charges are fixed

to the corners of a 0.30-m square, one to a corner, in such a way that the net force on any charge is directed toward the

center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

**27. A small spherical insulator of mass and charge is hung by a thread of negligible

mass. A charge of is held 0.150 m away from the sphere and directly to the right of it, so the thread makes

an angle with the vertical (see the drawing). Find

(a) the angle and

Answer:

REASONING The charged insulator experiences an electric force due to the

presence of the charged sphere shown in the drawing in the text. The forces acting on the

insulator are the downward force of gravity (i.e., its weight, ), the electrostatic

force (see Coulomb's law, Equation 18.1) pulling to the right, and the

tension in the thread pulling up and to the left at an angle with respect to the vertical,

as shown in the drawing in the problem statement. We can analyze the forces to determine

the desired quantities and .

SOLUTION

(a

)

We can see from the diagram given with the problem statement that

and

Dividing the first equation by the second yields

Solving for , we find that

(b

) Since , the tension can be obtained as follows:

(b) the tension in the thread.

Answer: 0.813 N

REASONING The charged insulator experiences an electric force due to the presence of the charged

sphere shown in the drawing in the text. The forces acting on the insulator are the downward force of gravity

(i.e., its weight, ), the electrostatic force (see Coulomb's law, Equation 18.1)

pulling to the right, and the tension in the thread pulling up and to the left at an angle with respect to the

vertical, as shown in the drawing in the problem statement. We can analyze the forces to determine the desired

quantities and .

SOLUTION

(a) We can see from the diagram given with the problem statement that

and

Dividing the first equation by the second yields

Solving for , we find that

(b

) Since , the tension can be obtained as follows:

REASONING The charged insulator experiences an electric force due to the presence of the charged sphere

shown in the drawing in the text. The forces acting on the insulator are the downward force of gravity (i.e., its weight,

), the electrostatic force (see Coulomb's law, Equation 18.1) pulling to the right, and the

tension in the thread pulling up and to the left at an angle with respect to the vertical, as shown in the drawing in the

problem statement. We can analyze the forces to determine the desired quantities and .

SOLUTION

(a) We can see from the diagram given with the problem statement that

and

Dividing the first equation by the second yields

Solving for , we find that

(b) Since , the tension can be obtained as follows:

**28. Two objects carry initial charges that are and , respectively, where . They are located 0.200 m

apart and behave like point charges. They attract each other with a force that has a magnitude of 1.20 N. The

objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial

positions. Now it is found that the objects repel one another with a force whose magnitude is equal to the

magnitude of the initial attractive force. What are the magnitudes of the initial charges on the objects?

Section 18.6 The Electric Field

Section 18.7 Electric Field Lines Section 18.8 The Electric Field Inside a Conductor: Shielding

29. At a distance from a point charge, the magnitude of the electric field created by the charge is 248 N/C. At a

distance from the charge, the field has a magnitude of 132 N/C. Find the ratio / .

Answer: 1.37

REASONING The electric field created by a point charge is inversely proportional to the square of the

distance from the charge, according to Equation 18.3. Therefore, we expect the distance to be greater than the

distance , since the field is smaller at than it is at . The ratio , then, should be greater than one.

SOLUTION Applying Equation 18.3 to each position relative to the charge, we have

Dividing the expression for by the expression for gives

Solving for the ratio gives

As expected, this ratio is greater than one.

30. Suppose you want to determine the electric field in a certain region of space. You have a small object of known

charge and an instrument that measures the magnitude and direction of the force exerted on the object by the electric

field.

(a) The object has a charge of and the instrument indicates that the electric force exerted on it is

, due east. What are the magnitude and direction of the electric field?

(b) What are the magnitude and direction of the electric field if the object has a charge of and the

instrument indicates that the force is , due west?

31. An electric field of 260 000 N/C points due west at a certain spot. What are the magnitude and direction of the force

that acts on a charge of at this spot?

Answer: 1.8 N due east

32. Review the important features of electric field lines discussed in Conceptual Example 13. Three point charges (

, and ) are at the corners of an equilateral triangle. Sketch in six electric field lines between the three

charges.

33. Four point charges have the same magnitude of and are fixed to the corners of a square that is 4.0 cm

on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that

exists at the center of the square.

Answer: 54 N/C

34. The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the

same distance of 6.1 cm from the origin O. For each of the situations in the drawing, determine the magnitude of the

net electric field at the origin.

35. A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and

is directed in the positive x direction. A point charge is placed at the origin. Determine the magnitude of the net electric field at

(a) ,

Answer: 7700 N/C

(b) , and

Answer: 1300 N/C

(c) .

Answer: 5500 N/C

36.

The membrane surrounding a living cell consists of aninner and an outer wall that are separated by a small

space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner

and outer walls has a magnitude of .

(a) What is the magnitude of the electric field within the cell membrane?

(b) Find the magnitude of the electric force that would be exerted on a potassium ion

placed inside the membrane.

37. A long, thin rod lies along the x axis, with its midpoint at the origin. In a vacuum, a

point charge is fixed to one end of the rod, and a point charge is fixed to the other end. Everywhere in the x,

y plane there is a constant external electric field that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.

Answer:

REASONING The drawing at the right shows the set-up. Here, the electric field E points along the axis

and applies a force of to the charge and a force of to the charge, where denotes the

magnitude of each charge. Each force has the same magnitude of , according to Equation 18.2. The torque is

measured as discussed in Section 9.1. According to Equation 9.1, the torque produced by each force has a magnitude

given by the magnitude of the force times the lever arm, which is the perpendicular distance between the point of

application of the force and the axis of rotation. In the drawing the axis is the axis of rotation and is midway between

the ends of the rod. Thus, the lever arm for each force is half the length of the rod or , and the magnitude of the

torque produced by each force is .

SOLUTION The and the force each cause the rod to rotate in the same sense about the axis. Therefore, the

torques from these forces reinforce one another. Using the expression for the magnitude of each torque,

we find that the magnitude of the net torque is

38. A point charge is placed in an external uniform electric field that has a magnitude of .

At what distance from the charge is the net electric field zero?

39. A tiny ball carries a charge of . What electric field (magnitude and direction) is

needed to cause the ball to float above the ground?

Answer:

directed downward

REASONING Two forces act on the charged ball (charge ); they are the downward force of gravity and

the electric force F due to the presence of the charge in the electric field E. In order for the ball to float, these two

forces must be equal in magnitude and opposite in direction, so that the net force on the ball is zero (Newton's second

law). Therefore, F must point upward, which we will take as the positive direction. According to Equation 18.2,

. Since the charge is negative, the electric field E must point downward, as the product in the expression

must be positive, since the force F points upward. The magnitudes of the two forces must be equal, so that

. This expression can be solved for .

SOLUTION The magnitude of the electric field E is

As discussed in the reasoning, this electric field points .

40. A proton and an electron are moving due east in a constant electric field that also points due east. The electric

field has a magnitude of . Determine the magnitude of the acceleration of the proton and the

electron.

41. Review Conceptual Example 12 before attempting to work this problem. The magnitude of each of the charges in

Figure 18.21 is . The lengths of the sides of the rectangles are 3.00 cm and 5.00 cm. Find the

magnitude of the electric field at the center of the rectangle in Figures 18.21a and b.

Answer:

in Figure 18.21a and in Figure 18.21b

42. Two charges are placed between the plates of a parallel plate capacitor. One charge is and the other is

. The charge per unit area on each of the plates has a magnitude of . The

magnitude of the force on due to equals the magnitude of the force on due to the electric field of the parallel

plate capacitor. What is the distance r between the two charges?

*43. A small object has a mass of and a charge of . It is placed at a certain spot where there

is an electric field. When released, the object experiences an acceleration of in the direction of the

axis. Determine the magnitude and direction of the electric field.

Answer:

directed along the

*44. A spring with an unstrained length of 0.074 m and a spring constant of 2.4 N/m hangs vertically downward

from the ceiling. A uniform electric field directed vertically upward fills the region containing the spring. A sphere

with a mass of and a net charge of is attached to the lower end of the spring. The spring is

released slowly, until it reaches equilibrium. The equilibrium length of the spring is 0.059 m. What is the magnitude

of the external electric field?

*45. Two point charges are located along the x axis: at , and at

. Two other charges are located on the y axis: at , and

at . Find the net electric field (magnitude and direction) at the origin.

Answer:

directed along the

REASONING The two charges lying on the axis produce no net electric field at the coordinate origin. This is

because they have identical charges, are located the same distance from the origin, and produce electric fields that point

in opposite directions. The electric field produced by at the origin points away from the charge, or along the

direction. The electric field produced by at the origin points toward the charge, or along the direction. The net

electric field is, then, , where and can be determined by using Equation 18.3.

SOLUTION The net electric field at the origin is

The plus sign indicates that .

*46. The total electric field consists of the vector sum of two parts. One part has a magnitude of

and points at an angle above the axis. The other part has a magnitude of and points

at an angle above the axis. Find the magnitude and direction of the total field. Specify the directional

angle relative to the axis.

*47. In Multiple-Concept Example 9 you can see the concepts that are important in this problem. A particle of

charge and mass is released from rest in a region where there is a constant electric field of

. What is the displacement of the particle after a time of ?

Answer:

*48. The drawing shows a positive point charge , a second point charge that may be positive or negative, and a

spot labeled P, all on the same straight line. The distance d between the two charges is the same as the distance

between and the spot P. With present, the magnitude of the net electric field at P is twice what it is when is

present alone. Given that , determine when it is

(a) positive and

(b) negative.

*49. Multiple-Concept Example 9 illustrates the concepts in this problem. An electron is released from rest at the negative

plate of a parallel plate capacitor. The charge per unit area on each plate is , and the plate

separation is . How fast is the electron moving just before it reaches the positive plate?

Answer:

**50. Two particles are in a uniform electric field that points in the direction and has a magnitude of 2500 N/C. The

mass and charge of particle 1 are and , while the corresponding values for

particle 2 are and . Initially the particles are at rest. The particles are both

located on the same electric field line but are separated from each other by a distance d. Particle 1 is located to the left

of particle 2. When released, they accelerate but always remain at this same distance from each other. Find d. **51. Two point charges of the same magnitude but opposite signs are fixed to either end of the base of an isosceles

triangle, as the drawing shows. The electric field at the midpoint M between the charges has a magnitude . The

field directly above the midpoint at point P has a magnitude . The ratio of these two field magnitudes is

. Find the angle in the drawing.

Answer:

REASONING AND SOLUTION The net electric field at point in Figure 1 is the vector sum of the fields

and , which are due, respectively, to the charges and . These fields are shown in Figure 2.

Figure 1

Figure 2

According to Equation 18.3, the magnitudes of the fields and are the same, since the triangle is an isosceles

triangle with equal sides of length . Therefore, . The vertical components of these two fields

cancel, while the horizontal components reinforce, leading to a total field at point that is horizontal and has a

magnitude of

At point in Figure 1, both and are horizontal and point to the right. Again using Equation 18.3, we find

Since , we have

But from Figure 1, we can see that . Thus, it follows that

The value for is, then, .

**52. The drawing shows an electron entering the lower left side of a parallel plate capacitor and exiting at the upper right

side. The initial speed of the electron is . The capacitor is 2.00 cm long, and its plates are separated

by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

**53. A small plastic ball with a mass of and with a charge of is suspended from an

insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with

the thread making an angle of with respect to the vertical. The area of each plate is . What is

the magnitude of the charge on each plate?

Answer:

Section

18.9 Gauss' Law

54. A spherical surface completely surrounds a collection of charges. Find the electric flux through the surface if the

collection consists of

(a) a single charge,

(b) a single charge, and

(c) both of the charges in (a) and (b).

55. The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface

1 has an area of , while surface 2 has an area of . The electric field in the drawing is uniform and has a

magnitude of 250 N/C. Find the magnitude of the electric flux through

(a) surface 1 and

Answer:

REASONING As discussed in Section 18.9, the magnitude of the

electric flux through a surface is equal to the magnitude of the component of

the electric field that is normal to the surface multiplied by the area of the

surface, ,where is the magnitude of the component of E that is

normal to the surface of area . We can use this expression and the figure in the

text to determine the desired quantities.

SOLUTION

(a

)

The magnitude of the flux through surface 1 is

(b

)

Similarly, the magnitude of the flux through surface 2 is

(b) surface 2.

Answer:

REASONING As discussed in Section 18.9, the magnitude of the electric flux through a surface is

equal to the magnitude of the component of the electric field that is normal to the surface multiplied by the

area of the surface, ,where is the magnitude of the component of E that is normal to the

surface of area . We can use this expression and the figure in the text to determine the desired quantities.

SOLUTION

(a) The magnitude of the flux through surface 1 is

(b) Similarly, the magnitude of the flux through surface 2 is

REASONING As discussed in Section 18.9, the magnitude of the electric flux through a surface is equal to

the magnitude of the component of the electric field that is normal to the surface multiplied by the area of the surface,

,where is the magnitude of the component of E that is normal to the surface of area . We can use this

expression and the figure in the text to determine the desired quantities.

SOLUTION

(a) The magnitude of the flux through surface 1 is

(b) Similarly, the magnitude of the flux through surface 2 is

56. A surface completely surrounds a charge. Find the electric flux through this surface when the surface

is

(a) a sphere with a radius of 0.50 m,

(b) a sphere with a radius of 0.25 m, and

(c) a cube with edges that are 0.25 m long.

57. A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude

. The magnitude of the electric flux through the surface is . What is the angle (less

than 90) between the direction of the electric field and the normal to the surface?

Answer:

58.

A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is:

, , , ,

, and . What is Q? *59. A solid nonconducting sphere has a positive charge q spread uniformly throughout its volume. The charge

density or charge per unit volume, therefore, is . Use Gauss' law to show that the electric field at a point

within the sphere at a radius r has a magnitude of . (Hint: For a Gaussian surface, use a sphere of radius r

centered within the solid sphere of radius. Note that the net charge within any volume is the charge density times the

volume.)

Answer: The answer is a proof.

*60. Two spherical shells have a common center. A charge is spread uniformly over the inner

shell, which has a radius of 0.050 m. A charge is spread uniformly over the outer shell, which has a

radius of 0.15 m. Find the magnitude and direction of the electric field at a distance (measured from the common

center) of

(a) 0.20 m,

(b) 0.10 m, and

(c) 0.025 m.

*61. A cube is located with one corner situated at the origin of an x, y, z coordinate system. One of the cube's faces lies

in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and

points in the direction of the axis. The magnitude of the field is 1500 N/C.

(a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces.

Answer:

The flux through the face in the x, z plane at is . The flux through the face

parallel to the x, z plane at is . The flux through each of the remaining four faces is zero.

REASONING The electric flux through each face of the cube is given by (see

Section 18.9) where is the magnitude of the electric field at the face, is the area of the face, and is the

angle between the electric field and the outward normal of that face. We can use this expression to calculate

the electric flux through each of the six faces of the cube.

SOLUTION

(a

) On the bottom face of the cube, the outward normal points parallel to the axis, in the opposite

direction to the electric field, and . Therefore,

On the top face of the cube, the outward normal points parallel to the axis, and . The

electric flux is, therefore,

On each of the other four faces, the outward normals are perpendicular to the direction of the electric

field, so . So for each of the four side faces,

(b

)

The total flux through the cube is

Therefore,

(b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as

Gauss' law predicts, since there is no net charge within the cube.

Answer:

REASONING The electric flux through each face of the cube is given by (see

Section 18.9) where is the magnitude of the electric field at the face, is the area of the face, and is the

angle between the electric field and the outward normal of that face. We can use this expression to calculate

the electric flux through each of the six faces of the cube.

SOLUTION

(a

) On the bottom face of the cube, the outward normal points parallel to the axis, in the opposite

direction to the electric field, and . Therefore,

On the top face of the cube, the outward normal points parallel to the axis, and . The

electric flux is, therefore,

On each of the other four faces, the outward normals are perpendicular to the direction of the electric

field, so . So for each of the four side faces,

(b

)

The total flux through the cube is

Therefore,

REASONING The electric flux through each face of the cube is given by (see Section

18.9) where is the magnitude of the electric field at the face, is the area of the face, and is the angle between the

electric field and the outward normal of that face. We can use this expression to calculate the electric flux through

each of the six faces of the cube.

SOLUTION

(a) On the bottom face of the cube, the outward normal points parallel to the axis, in the opposite direction to the

electric field, and . Therefore,

On the top face of the cube, the outward normal points parallel to the axis, and . The electric flux

is, therefore,

On each of the other four faces, the outward normals are perpendicular to the direction of the electric field, so

. So for each of the four side faces,

(b

)

The total flux through the cube is

Therefore,

**62. A long, thin, straight wire of length L has a positive charge Q distributed uniformly along it. Use Gauss' law to show

that the electric field created by this wire at a radial distance r has a magnitude of , where

. (Hint: For a Gaussian surface, use a cylinder aligned with its axis along the wire and note that the cylinder has a flat surface at either end, as well as a curved surface.)

Copyright 2012 John Wiley & Sons, Inc. All rights reserved.

Problems

Section 19.1 Potential Energy Section 19.2 The Electric Potential Difference

1. During a particular thunderstorm, the electric potential difference between a cloud and the ground is

, with the cloud being at the higher potential. What is the change in an electron's

electric potential energy when the electron moves from the ground to the cloud?

Answer:

2.

A particle with a charge of and a mass of is released from rest at point A and accelerates

toward point B, arriving there with a speed of 42 m/s. The only force acting on the particle is the electric force.

(a) Which point is at the higher potential? Give your reasoning.

(b) What is the potential difference between A and B?

3.

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How

much work is done by the electric force when a sodium ion moves from the outside to the inside?

Answer:

REASONING AND SOLUTION Combining Equations 19.1 and 19.3, we have

4. A particle has a charge of and moves from point A to point B, a distance of 0.20 m. The particle

experiences a constant electric force, and its motion is along the line of action of the force. The difference between the

particle's electric potential energy at A and at B is .

(a) Find the magnitude and direction of the electric force that acts on the particle.

(b) Find the magnitude and direction of the electric field that the particle experiences.

5. Multiple-Concept Example 3 employs some of the concepts that are needed here. An electric car accelerates for

7.0 s by drawing energy from its 290-V battery pack. During this time, 1200 C of charge passes through the battery

pack. Find the minimum horsepower rating of the car.

Answer: 67 hp

6. Review Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube,

electrons strike the screen after being accelerated from rest through a potential difference of 25 000 V. The speeds of

the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken

into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

7. Multiple-Concept Example 4 deals with the concepts that are important in this problem. As illustrated in Figure

19.5b, a negatively charged particle is released from rest at point B and accelerates until it reaches point A. The mass

and charge of the particle are and , respectively. Only the gravitational force and the

electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at

A is 36 V greater than that at B; in other words, . What is the translational speed of the particle at point A?

Answer: 19 m/s

REASONING The translational speed of the particle is related to the particle's translational kinetic energy,

which forms one part of the total mechanical energy that the particle has. The total mechanical energy is conserved,

because only the gravitational force and an electrostatic force, both of which are conservative forces, act on the particle

(see Section 6.5). Thus, we will determine the speed at point by utilizing the principle of conservation of mechanical

energy.

SOLUTION The particle's total mechanical energy is

Since the particle does not rotate the angular speed is always zero, and since there is no elastic force we may omit the

terms and from this expression. With this in mind, we express the fact that (energy is conserved)

as follows:

This equation can be simplified further, since the particle travels horizontally, so that , with the result that

Solving for gives

According to Equation 19.4, the difference in electric potential energies is related to the electric

potential difference :

Substituting this expression into the expression for gives

8. An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same

magnitude. In the process, the electron acquires a speed , while the proton acquires a speed . Find the ratio

.

*9. The potential at location A is 452 V. A positively charged particle is released there from rest and arrives at

location B with a speed . The potential at location C is 791 V, and when released from rest from this spot, the

particle arrives at B with twice the speed it previously had, or . Find the potential at B.

Answer: 339 V

REASONING The only force acting on the moving charge is the conservative electric force. Therefore, the

total energy of the charge remains constant. Applying the principle of conservation of energy between locations A and

B, we obtain

Since the charged particle starts from rest, . The difference in potential energies is related to the difference in

potentials by Equation 19.4, . Thus, we have

(1)

Similarly, applying the conservation of energy between locations C and B gives

(2)

Dividing Equation 1 by Equation 2 yields

This expression can be solved for .

SOLUTION Solving for , we find that

*10. A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060

eV as the particle moves from position A to position B. The electric potential at A is , and the electric

potential at B is . Determine the charge of the particle. Include the algebraic sign ( or ) with your

answer.

*11. During a lightning flash, there exists a potential difference of between

a cloud and the ground. As a result, a charge of is transferred from the ground to the cloud.

(a) How much work is done on the charge by the electric force?

Answer:

(b) If the work done by the electric force were used to accelerate a 1100-kg automobile from rest, what

would be its final speed?

Answer:

(c) If the work done by the electric force were converted into heat, how many kilograms of water at

could be heated to ?

Answer:

**12. A particle is uncharged and is thrown vertically upward from ground level with a speed of 25.0 m/s. As a result,

it attains a maximum height h. The particle is then given a positive charge and reaches the same maximum

height h when thrown vertically upward with a speed of 30.0 m/s. The electric potential at the height h exceeds the electric potential at ground level. Finally, the particle is given a negative charge . Ignoring air

resistance, determine the speed with which the negatively charged particle must be thrown vertically upward,

so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity. Section

19.3 The Electric

Potential Difference Created by Point Charges

13. Two point charges, and , are separated by 1.20 m. What is the electric potential midway

between them?

Answer:

14. An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought

together to form a hydrogen atom, in which the electron orbits the proton at an average distance of .

What is , which is the change in the electric potential energy?

15. Two charges A and B are fixed in place, at different distances from a certain spot. At this spot the potentials due

to the two charges are equal. Charge A is 0.18 m from the spot, while charge B is 0.43 m from it. Find the ratio

of the charges.

Answer: 2.4

REASONING The potential of each charge at a distance away is given by Equation 19.6 as . By

applying this expression to each charge, we will be able to find the desired ratio, because the distances are given for

each charge.

SOLUTION According to Equation 19.6, the potentials of each charge are

Since we know that , it follows that

16. The drawing shows a square, each side of which has a length of . On two corners of the square are

fixed different positive charges, and . Find the electric potential energy of a third charge

placed at corner A and then at corner B.

17. The drawing shows four point charges. The value of q is , and the distance d is 0.96 m. Find the total

potential at the location P. Assume that the potential of a point charge is zero at infinity.

Probolem 17

Answer:

REASONING The electric potential at a distance from a point charge is given by Equation 19.6 as

. The total electric potential at location due to the four point charges is the algebraic sum of the

individual potentials.

SOLUTION The total electric potential at is (see the drawing)

Substituting in the numbers gives

18. A charge of is fixed at the center of a square that is 0.64 m on a side. How much work is done by the

electric force as a charge of is moved from one corner of the square to any other empty corner? Explain.

19. The drawing shows six point charges arranged in a rectangle. The value of q is , and the distance d is 0.13 m.

Find the total electric potential at location P, which is at the center of the rectangle.

Answer:

20. Location A is 3.00 m to the right of a point charge q. Location B lies on the same line and is 4.00 m to the right of the

charge. The potential difference between the two locations is . What are the magnitude and sign

of the charge?

21. Identical charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign)

should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is 0 V?

Answer:

22.

Charges of and are fixed in place, with a distance of 2.00 m between them. A dashed line is drawn

through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance L from the negative charge, there is at least one spot where the total potential is zero. Find L.

*23. Determine the electric potential energy for the array of three charges in the drawing, relative to its value when

the charges are infinitely far away and infinitely far apart.

Probolem 23

Answer:

REASONING Initially, the three charges are infinitely far apart. We will proceed as in Example 8 by adding

charges to the triangle, one at a time, and determining the electric potential energy at each step. According to Equation

19.3, the electric potential energy EPE is the product of the charge and the electric potential at the spot where the

charge is placed, . The total electric potential energy of the group is the sum of the energies of each step in

assembling the group.

SOLUTION Let the corners of the triangle be numbered clockwise as 1, 2 and 3, starting with the top corner. When

the first charge is placed at a corner 1, the charge has no electric potential energy, . This

is because the electric potential produced by the other two charges at corner 1 is zero, since they are infinitely far

away.

Once the 8.00- charge is in place, the electric potential that it creates at corner 2 is

where is the distance between corners 1 and 2, and . When the 20.0- charge is

placed at corner 2, its electric potential energy is

The electric potential at the remaining empty corner is the sum of the potentials due to the two charges that are

already in place on corners 1 and 2:

where , , , and . When the third charge is

placed at corner 3, its electric potential energy is

The electric potential energy of the entire array is given by

*24.

Two identical point charges are fixed at diagonally opposite corners of a square with

sides of length 0.480 m. A test charge , with a mass of , is released from

rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the

square.

*25. Two protons are moving directly toward one another. When they are very far apart, their initial speeds are

. What is the distance of closest approach?

Answer:

*26.

Four identical charges ( each) are brought from infinity and fixed to a straight line. The charges are

located 0.40 m apart. Determine the electric potential energy of this group.

*27. A charge of is fixed in place. From a horizontal distance of 0.0450 m, a particle of mass

and charge is fired with an initial speed of 65.0 m/s directly toward the fixed charge. How

far does the particle travel before its speed is zero?

Answer: 0.0342 m

REASONING The only force acting on the moving charge is the conservative electric force. Therefore, the

sum of the kinetic energy KE and the electric potential energy EPE is the same at points A and B:

Since the particle comes to rest at B, . Combining Equations 19.3 and 19.6, we have

and

where is the initial distance between the fixed charge and the moving charged particle, and is the distance between

the charged particles after the moving charge has stopped. Therefore, the expression for the conservation of energy

becomes

This expression can be solved for . Once is known, the distance that the charged particle moves can be determined.

SOLUTION Solving the expression above for gives

Therefore, the charge moves a distance of .

*28. Identical point charges of are fixed to diagonally opposite corners of a square. A third charge is then

fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without

changing magnitudes. Find the sign and magnitude of the third charge.

**29. One particle has a mass of and a charge of . A second particle has a mass of

and the same charge. The two particles are initially held in place and then released. The particles fly

apart, and when the separation between them is 0.100 m, the speed of the particle is 125 m/s. Find

the initial separation between the particles.

Answer:

**30. Review Conceptual Example 7 as background for this problem. A positive charge is located to the left of a

negative charge . On a line passing through the two charges, there are two places where the total potential is zero.

The first place is between the charges and is 4.00 cm to the left of the negative charge. The second place is 7.00 cm to

the right of the negative charge.

(a) What is the distance between the charges?

(b) Find , the ratio of the magnitudes of the charges.

Section 19.4 Equipotential Surfaces andTheir Relation to the Electric Field 31.

Two equipotential surfaces surround a point charge. How far is the 190-V surface from the 75.0-V

surface?

Answer: 1.1 m

32. An equipotential surface that surrounds a point charge q has a potential of 490 V and an area of . Determine q.

33.

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively.

Because of these charges, a potential difference of about 0.070 V exists across the membrane. The thickness of the

cell membrane is . What is the magnitude of the electric field in the membrane?

Answer:

REASONING AND SOLUTION From Equation 19.7a we know that , where is the potential

difference between the two surfaces of the membrane, and is the distance between them. If is a point on the

positive surface and is a point on the negative surface, then . The electric field

between the surfaces is

34.

A positive point charge is surrounded by an equipotential surface A, which has a radius

of . A positive test charge moves from surface A to another equipotential

surface B, which has a radius . The work done as the test charge moves from surface A to surface B is

. Find .

35. A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.75

mm. When an electric spark jumps between them, the magnitude of the electric field is . What is the

magnitude of the potential difference between the conductors?

Answer:

REASONING The magnitude of the electric field is given by Equation 19.7a (without the minus sign) as

, where is the potential difference between the two metal conductors of the spark plug, and is the

distance between the two conductors. We can use this relation to find .

SOLUTION The potential difference between the conductors is

36. The drawing that accompanies Problem 60 shows a graph of a set of equipotential surfaces in cross section. The grid

lines are 2.0 cm apart. Determine the magnitude and direction of the electric field at position D. Specify whether the electric field points toward the top or the bottom of the drawing.

*37. An electric field has a constant value of and is directed downward. The field is the same

everywhere. The potential at a point P within this region is 155 V. Find the potential at the following points:

(a) directly above P,

Answer: 179 V

REASONING The drawing shows the electric field and the three points, , , and , in the

vicinity of point , which we take as the origin. We choose the upward direction as being positive. Thus,

, since the electric field points straight down. The electric potential at points and

can be determined from Equation 19.7a as , since and are known. Since the path from to

is perpendicular to the electric field, no work is done in moving a charge along such a path. Thus, the

potential difference between these two points is zero.

SOLUTION

(a) The potential difference between points and is . The potential at is

(b) The potential difference between points and is . The potential at is

(c) Since the path from to is perpendicular to the electric field and no work is done in moving a

charge along such a path, it follows that . Therefore, .

(b) directly below P,

Answer: 143 V

REASONING The drawing shows the electric field and the three points, , , and , in the

vicinity of point , which we take as the origin. We choose the upward direction as being positive. Thus,

, since the electric field points straight down. The electric potential at points and

can be determined from Equation 19.7a as , since and are known. Since the path from to

is perpendicular to the electric field, no work is done in moving a charge along such a path. Thus, the

potential difference between these two points is zero.

SOLUTION

(a) The potential difference between points and is . The potential at is

(b) The potential difference between points and is . The potential at is

(c) Since the path from to is perpendicular to the electric field and no work is done in moving a

charge along such a path, it follows that . Therefore, .

(c) directly to the right of P.

Answer: 155 V

REASONING The drawing shows the electric field and the three points, , , and , in the

vicinity of point , which we take as the origin. We choose the upward direction as being positive. Thus,

, since the electric field points straight down. The electric potential at points and

can be determined from Equation 19.7a as , since and are known. Since the path from to

is perpendicular to the electric field, no work is done in moving a charge along such a path. Thus, the

potential difference between these two points is zero.

SOLUTION

(a) The potential difference between points and is . The potential at is

(b) The potential difference between points and is . The potential at is

(c) Since the path from to is perpendicular to the electric field and no work is done in moving a

charge along such a path, it follows that . Therefore, .

REASONING The drawing shows the electric field and the three points, , , and , in the vicinity of

point , which we take as the origin. We choose the upward direction as being positive. Thus,

, since the electric field points straight down. The electric potential at points and can be

determined from Equation 19.7a as , since and are known. Since the path from to is

perpendicular to the electric field, no work is done in moving a charge along such a path. Thus, the potential

difference between these two points is zero.

SOLUTION

(a) The potential difference between points and is . The potential at is

(b) The potential difference between points and is . The potential at is

(c) Since the path from to is perpendicular to the electric field and no work is done in moving a charge along

such a path, it follows that . Therefore, .

*38. An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive

plate (see the drawing). The plates are separated by a distance of 1.2 cm, and the electric field within the capacitor has

a magnitude of . What is the kinetic energy of the electron just as it reaches the positive plate?

*39. The drawing shows the electric potential as a function of distance along the x axis. Determine the magnitude of

the electric field in the region

(a) A to B,

Answer: 0 V/m

(b) B to C, and

Answer:

(c) C to D.

Answer: 5.0 V/m

*40. At a distance of 1.60 m from a point charge of , there is an equipotential surface. At greater distances

there are additional equipotential surfaces. The potential difference between any two successive surfaces is

. Starting at a distance of 1.60 m and moving radially outward, how many of the additional

equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not

include the starting surface.

*41. The drawing shows a uniform electric field that points in the negative y direction; the magnitude of the

field is 3600 N/C. Determine the electric potential difference

Probolem 41

(a) between points A and B,

Answer:

0 V

(b) between points B and C, and

Answer:

(c) between points C and A.

Answer:

Section

19.5 Capacitors and Dielectrics

42. What is the capacitance of a capacitor that stores of charge on its plates when a voltage of 1.5 V is applied

between them?

43.

The electric potential energy stored in the capacitor of a defibrillator is 73 J, and the capacitance is

. What is the potential difference that exists across the capacitor plates?

Answer:

REASONING According to Equation 19.11b, the energy stored in a capacitor with capacitance and

potential across its plates is .

SOLUTION Therefore, solving Equation 19.11b for , we have

44. Two identical capacitors store different amounts of energy: capacitor A stores , and capacitor B stores

. The voltage across the plates of capacitor B is 12 V. Find the voltage across the plates of capacitor A.

45. The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the

flash. In one such unit, the potential difference between the plates of an capacitor is 280 V.

(a) Determine the energy that is used to produce the flash in this unit.

Answer: 33 J

(b) Assuming that the flash lasts for , find the effective power or wattage of the flash. Answer: 8500 W

46. The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this

voltage causes the capacitor to store of charge and of energy. When used with capacitor B, which

has a capacitance of , this voltage causes the capacitor to store a charge that has a magnitude of . Determine

.

47. A parallel plate capacitor has a capacitance of when filled with a dielectric. The area of each plate is

and the separation between the plates is . What is the dielectric constant of the dielectric?

Answer: 5.3

REASONING AND SOLUTION Equation 19.10 gives the capacitance for a parallel plate capacitor filled

with a dielectric of constant : . Solving for , we have

48. Two capacitors are identical, except that one is empty and the other is filled with a dielectric . The

empty capacitor is connected to a 12.0-V battery. What must be the potential difference across the plates of the

capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?

49.

The membrane that surrounds a certain type of living cell has a surface area of and a thickness

of . Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of

5.0.

(a) The potential on the outer surface of the membrane is greater than that on the inside surface.

How much charge resides on the outer surface?

Answer:

(b) If the charge in part (a) is due to positive ions (charge ), how many such ions are present on the outer

surface?

Answer:

*50.

Capacitor A and capacitor B both have the same voltage across their plates. However, the energy of capacitor A

can melt m kilograms of ice at , while the energy of capacitor B can boil away the same amount of water at

. The capacitance of capacitor A is . What is the capacitance of capacitor B?

*51. What is the potential difference between the plates of a 3.3-F capacitor that stores sufficient energy to

operate a 75-W light bulb for one minute?

Answer: 52 V

REASONING According to Equation 19.11b, the energy stored in a capacitor with a capacitance and

potential across its plates is . Once we determine how much energy is required to operate a 75-W

light bulb for one minute, we can then use the expression for the energy to solve for .

SOLUTION The energy stored in the capacitor, which is equal to the energy required to operate a 75-W bulb for one

minute , is

Therefore, solving Equation 19.11b for , we have

*52. A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical

shell is , and that of the outer shell is . When the cylinders carry equal and opposite

charges of magnitude , the electric field between the plates has an average magnitude of

and is directed radially outward from the inner shell to the outer shell. Determine

(a) the magnitude of the potential difference between the cylindrical shells and

(b) the capacitance of this capacitor.

*53. Review Conceptual Example 11 before attempting this problem. An empty capacitor is connected to a 12.0-V battery

and charged up. The capacitor is then disconnected from the battery, and a slab of dielectric material is

inserted between the plates. Find the amount by which the potential difference across the plates changes. Specify

whether the change is an increase or a decrease.

Answer: 7.7 V decrease

*54. An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The

capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of

this change, what is the new voltage between the plates of the capacitor?

**55. The potential difference between the plates of a capacitor is 175 V. Midway between the plates, a proton and an

electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative

plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive

plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

Answer:

REASONING If we assume that the motion of the proton and the electron is horizontal in the direction,

the motion of the proton is determined by Equation 2.8, , where is the distance traveled by the

proton, is its initial speed, and is its acceleration. If the distance between the capacitor places is , then this

relation becomes , or

(1)

We can solve Equation 1 for the initial speed of the proton, but, first, we must determine the time and the

acceleration of the proton . Since the proton strikes the negative plate at the same instant the electron strikes the

positive plate, we can use the motion of the electron to determine the time .

For the electron, , where we have taken into account the fact that the electron is released from rest.

Solving this expression for we have . Substituting this expression into Equation 1, we have

(2)

The accelerations can be found by noting that the magnitudes of the forces on the electron and proton are equal, since

these particles have the same magnitude of charge. The force on the electron is , and the

acceleration of the electron is, therefore,

(3)

Newton's second law requires that , so that

(4)

Combining Equations 2, 3 and 4 leads to the following expression for , the initial speed of the proton:

SOLUTION Substituting values into the expression above, we find

**56. If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric

will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude that

the electric field can have without breakdown occurring. The dielectric strength of air is , and that of

neoprene rubber is . A certain air-gap, parallel plate capacitor can store no more than 0.075 J of

electrical energy before breaking down. How much energy can this capacitor store without breaking down after the

gap between its plates is filled with neoprene rubber?

Copyright 2012 John Wiley & Sons, Inc. All rights reserved.

Problems

Section 20.1 Electromotive Force and Current Section 20.2 Ohm's Law

1.

A defibrillator is used during a heart attack to restore the heart to its normal beating pattern (see Section 19.5).

A defibrillator passes of current through the torso of a person in 2.0 ms.

(a) How much charge moves during this time?

Answer:

(b) How many electrons pass through the wires connected to the patient?

Answer:

electrons

2. An especially violent lightning bolt has an average current of lasting 0.138 s. How much charge is

delivered to the ground by the lightning bolt?

3. A battery charger is connected to a dead battery and delivers a current of for 5.0 hours, keeping the

voltage across the battery terminals at in the process. How much energy is delivered to the battery?

Answer:

REASONING AND SOLUTION First determine the total charge delivered to the battery using Equation

20.1:

To find the energy delivered to the battery, multiply this charge by the energy per unit charge (i.e., the voltage) to get

4. A coffee-maker contains a heating element that has a resistance of . This heating element is energized by a

outlet. What is the current in the heating element?

5.

Suppose that the resistance between the walls of a biological cell is .

(a) What is the current when the potential difference between the walls is 75 mV?

Answer:

(b)

If the current is composed of ions , how many such ions flow in ?

Answer:

ions

*6. A car battery has a rating of 220 amperehours (Ah). This rating is one indication of the total charge that the battery can provide to a circuit before failing.

(a) What is the total charge (in coulombs) that this battery can provide?

(b) Determine the maximum current that the battery can provide for 38 minutes.

*7. A resistor is connected across the terminals of a battery, which delivers of energy to the

resistor in six hours. What is the resistance of the resistor?

Answer:

*8. The resistance of a bagel toaster is . To prepare a bagel, the toaster is operated for one minute from a

outlet. How much energy is delivered to the toaster?

**9. A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current

whose value is .

(a) How many protons strike the target in ?

Answer:

protons

REASONING The number of protons that strike the target is equal to the amount of

electric charge striking the target divided by the charge of a proton, . From

Equation 20.1, the amount of charge is equal to the product of the current and the time . We can

combine these two relations to find the number of protons that strike the target in 15 seconds.

The heat that must be supplied to change the temperature of the aluminum sample of mass by an

amount is given by Equation 12.4 as , where is the specific heat capacity of

aluminum. The heat is provided by the kinetic energy of the protons and is equal to the number of

protons that strike the target times the kinetic energy per proton. Using this reasoning, we can find

the change in temperature of the block for the 15 second-time interval.

SOLUTION

(a) The number of protons that strike the target is

(b) The amount of heat provided by the kinetic energy of the protons is

Since and since Table 12.2 gives the specific heat of aluminum as

, the change in temperature of the block is

(b) Each proton has a kinetic energy of . Suppose the target is a 15-gram block of

aluminum, and all the kinetic energy of the protons goes into heating it up. What is the change in

temperature of the block that results from the 15-s bombardment of protons?

Answer:

REASONING The number of protons that strike the target is equal to the amount of

electric charge striking the target divided by the charge of a proton, . From

Equation 20.1, the amount of charge is equal to the product of the current and the time . We can

combine these two relations to find the number of protons that strike the target in 15 seconds.

The heat that must be supplied to change the temperature of the aluminum sample of mass by an

amount is given by Equation 12.4 as , where is the specific heat capacity of

aluminum. The heat is provided by the kinetic energy of the protons and is equal to the number of

protons that strike the target times the kinetic energy per proton. Using this reasoning, we can find

the change in temperature of the block for the 15 second-time interval.

SOLUTION

(a) The number of protons that strike the target is

(b) The amount of heat provided by the kinetic energy of the protons is

Since and since Table 12.2 gives the specific heat of aluminum as

, the change in temperature of the block is

REASONING The number of protons that strike the target is equal to the amount of electric charge

striking the target divided by the charge of a proton, . From Equation 20.1, the amount of

charge is equal to the product of the current and the time . We can combine these two relations to find the

number of protons that strike the target in 15 seconds.

The heat that must be supplied to change the temperature of the aluminum sample of mass by an amount

is given by Equation 12.4 as , where is the specific heat capacity of aluminum. The heat is

provided by the kinetic energy of the protons and is equal to the number of protons that strike the target times

the kinetic energy per proton. Using this reasoning, we can find the change in temperature of the block for the

15 second-time interval.

SOLUTION

(a) The number of protons that strike the target is

(b) The amount of heat provided by the kinetic energy of the protons is

Since and since Table 12.2 gives the specific heat of aluminum as

, the change in temperature of the block is

Section 20.3 Resistance and Resistivity

10. The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows

three situations in which the current takes different paths through a piece of material. Each of the rectangular pieces is

made from a material whose resistivity is , and the unit of length in the drawing is

. Each piece of material is connected to a battery. Find

(a) the resistance and

(b) the current in each case.

11. Two wires are identical, except that one is aluminum and one is copper. The aluminum wire has a resistance of

. What is the resistance of the copper wire?

Answer:

12. A cylindrical wire has a length of and a radius of . It carries a current of , when a voltage of

is applied across the ends of the wire. From what material in Table 20.1 is the wire made?

13. A coil of wire has a resistance of at and at . What is the temperature coefficient of

resistivity?

Answer:

14. A large spool in an electrician's workshop has of insulation-coated wire coiled around it. When the electrician

connects a battery to the ends of the spooled wire, the resulting current is . Some weeks later, after cutting off

various lengths of wire for use in repairs, the electrician finds that the spooled wire carries a 3.1-A current when the

same battery is connected to it. What is the length of wire remaining on the spool?

15. Two wires have the same length and the same resistance. One is made from aluminum and the other from

c