In the previous chapter we discussed the field concept, and in particular electric fields. The picture of an electric field that emerged is that of a collection of arrows, one for each point in space, where each arrow describes the strength and direction of the electric field at a particular point in space.

In this chapter we'll discuss an alternative way to describe an electric field, using the concept of electric potential. I want to emphasize that we are still describing the same fundamental reality, the electric field, but doing so in two different ways. One way is by specifying the electric field itself (and its associated collection of arrows), and the other way is to specify a set of scalar quantities; that is, the electric potential at each point in space.

There are advantages and disadvantages to each perspective on electric fields; as usual, having both perspectives at our

Key concepts:

electric potential•electric potential energy•the electron-volt (eV), a convenient unit of energy when dealing with atomic or subatomic phenomena

•

a point charge○

a constant electric field○

electric potential formula for: •

connection between electric potential and electric field•properties of equipotential surfaces•properties of a conductor in electrostatic equilibrium•capacitance•capacitance of a parallel-plate capacitor•capacitance of a parallel-plate capacitor with a dielectric medium between the plates

•

energy stored in an electric field•

Chapter 19 Electric Potential Energy and Electric PotentialSunday, January 31, 2010 10:37 PM

Ch19L Page 1

on electric fields; as usual, having both perspectives at our fingertips, and being able to smoothly switch from one perspective to the other, is a great aid to our understanding and also our ability to solve problems.

To understand the concept of electric potential, and how it connects with the electric field concept, begin by thinking about topographic maps.

The class discussion about the topographic map led to the following conclusions:

points on the same contour line are at the same elevation; thus, contour lines are "equal-elevation" lines

•

the landscape is steepest where the contour lines are closest together; on the other hand, where the contour lines are far apart the landscape is not very steep

•

It's interesting to interpret the topographic map in terms of gravitational forces and gravitational potential energy, as we discussed in mechanics. Recall that the expression

Ch19L Page 2

as we discussed in mechanics. Recall that the expression for gravitational potential energy for an object of mass mat an elevation h is U = mgh, where g is nearly constant near the Earth's surface. It's useful to separate the gravitational potential energy into a product of two factors, as follows: U = (m)(gh), where the mass m "belongs" to the object, and the quantity gh "belongs" to the gravitational field. The quantity gh is called the gravitational potential, and is symbolized by V, so that V = gh. Make sure to carefully distinguish between the gravitational potential (which is specified at each point in space, and is independent of any object that happens to be located at a particular point in space), and the gravitational potential energy of an object of mass m (which depends on the mass of the object as well as its location in space).

Contour lines, which are lines of equal elevation, are also lines of equal gravitational potential, and so we could just as well call them equipotential lines, or equipotentials for short.

For simplicity, imagine that the landscape in the topographic map is nice and smooth (and let's say it's also frictionless, for simplicity). It's natural that an object that is released at some point in the landscape will slide downwards to lower elevations; thus, the gravitational force points in the direction of decreasing elevation; in other words, the gravitational force points in the direction of decreasing gravitational potential. Thus, in the absence of other forces, an object will naturally move in the direction of decreasing gravitational potential.

It's also worth noting that if an object is released from rest in our idealized landscape, the object will initially move perpendicular to the contour lines. Another way of saying this is that the gravitational force is perpendicular to the equipotential lines. (More accurately, the component of the gravitational force that points along the slope is perpendicular to the equipotential lines.)

Also note that the zero-level for elevation in the topographic map is arbitrary; similarly, the zero-level for gravitational potential is arbitrary. It's only differences in gravitational potential that are

Ch19L Page 3

arbitrary. It's only differences in gravitational potential that are physically meaningful, as we shall see in examples later in this chapter. Thus, we are free to select the zero-level for gravitational potential in whatever way is convenient.

By analogy, the same conclusions hold for electric potential, as we'll now describe.

The usual way of visualizing an electric field is as a collection of arrows, one arrow attached to each point in space. The length of the arrow attached to a certain point in space represents the magnitude of the electric field at that point, and the direction of the arrow represents the direction of the electric field at that point.

By analogy with topographic maps, the electric field can also be represented by a collection of equipotential surfaces; somewhat like contour lines, each of which has constant elevation, each equipotential surfaces has constant electric potential. At each point in space, the electric field vector is perpendicular to the equipotential surface passing through the same point. The electric field is strong where the equipotentials are close together, and the electric field is weak where the equipotentials are spaced far apart.

The zero-level for electric potential is arbitrary; it's only differences in potential that are physically meaningful. Thus, we are free to choose the zero-level for electric potential at our convenience.

By convention, electric field vectors point in the direction of decreasing potential. Because electric field vectors point away from a single positive charge, we can conclude that the electric potential decreases as you move away from a single positive charge. Similarly, electric field vectors point towards a single negative charge, so electric potential decreases as you move towards a single negative charge.

The unit of electric potential is the volt, symbol V, which is equivalent to a joule per coulomb: 1 V = 1 J/C.

Ch19L Page 4

The textbook uses the previous equation, but remember that this equation represents magnitudes only. You have to put the signs in by

The first row of formulas applies to a uniform gravitational field. The work done in moving an object a distance h against a uniform gravitational force at a constant speed is W = Fh = mgh, and this work appears as an increase in the gravitational potential energy. The second row applies to a uniform electric field. The work done in moving a charged particle with charge q a distance d against a uniform electric field is W = Fd = qEd, and this work appears as an increase in the electric potential energy qV. Equating these two expressions for the work and the increase in electric potential energy results in qEd = qV, and therefore Ed = V for a uniform electric field.

equivalent to a joule per coulomb: 1 V = 1 J/C.

Compare gravitational and electric quantities; the connection between the electric field and the electric potential is valid provided the electric field is uniform:

Ch19L Page 5

equation represents magnitudes only. You have to put the signs in by hand, by remembering that the electric field points in the direction of decreasing potential.

At least, this is what our previous textbook wrote. Our current textbook writes

So our current textbook explicitly put the negative sign in the formula. This again accurately reflects the convention that the electric field points in the direction of decreasing potential.

Ch19L Page 6

Some examples of electric field / electric potential

Potential pattern for a positive point charge:

Potential pattern for a negative point charge:

Ch19L Page 7

Potential pattern for a dipole:

Potential pattern for a positive point charge (left diagram), a dipole (centre diagram), and a parallel-plate capacitor (right diagram):

Ch19L Page 8

Potential pattern for a constant electric field (parallel-plate capacitor):

Remember that for a single positive point charge, the potential decreases as you move away from the positive point charge, because the electric field points away from the positive point charge. For a single negative point charge, the potential decreases as you move towards the negative point charge, because the electric field points towards the negative point charge.

Note: The unit for electric potential is the volt, V.

The unit of electric field is N/C, as we saw in the previous chapter. However, now that we have learned the connection between electric field and electric potential, we can see that another, equivalent unit for the electric field is V/m.

Ch19L Page 9

Example: Suppose you have a uniform (i.e., constant) electric field of magnitude 20 V/m, in the positive x-direction. Let's label the potential at x = 2 m as 0 V. Determine the electric potential at x = 0 m, x = 1 m, x = 3 m, and x = 4 m.

Solution: Remember that the electric field points in the direction of decreasing potential.

Ch19L Page 10

Details of alternative solution:

Ch19L Page 11

Note that positive work is done in moving the charged particle from A to B, and because the charge is positive, this means that the potential of B is greater than the potential of A. On the other hand, negative work is done in moving the charged particle from C to B, and because the charge is positive, this means that the potential of B is less than the potential of C. Thus,

Ch19L Page 12

Ch19L Page 13

Ch19L Page 14

Ch19L Page 15

Ch19L Page 16

2/4/2010, 10:12 AM

Ch19L Page 17

Ch19L Page 18

Ch19L Page 19

Ch19L Page 20

Ch19L Page 21

Ch19L Page 22

Ch19L Page 23

Ch19L Page 24

Ch19L Page 25

Ch19L Page 26

Ch19L Page 27

Ch19L Page 28

Example:

Ch19L Page 29

Example:

Ch19L Page 30

Selected Problems and Solutions

Problem: It takes 3.0 µ J of work to move a 15 nC charge from

point A to point B. It takes 5.0 µ J of work to move the same charge from C to B. What is the potential difference

VC VA?

Ch19L Page 31

Problem: A proton has been accelerated from rest

through a potential difference of 1000 V. Calculate its final kinetic energy in (a) electron volts, and (b) joules. (c.) Calculate its final speed.

Ch19L Page 32

Problem: What is the potential difference V34 in the figure?

Ch19L Page 33

Problem: At a distance r from a point charge, the electric

potential is 3000 V and the magnitude of the electric field is 2.0 × 105 V/m. (a) Calculate the distance r. (b) Calculate the electric potential and the magnitude of the electric

Ch19L Page 34

the electric potential and the magnitude of the electric field at a distance r/2 from the point charge.

Ch19L Page 35

Problem: A +3.0 nC charge is at x = 0 cm and a 1.0 nC charge is at x = 4 cm. At which point or points along the x-axis is the electric potential zero?

Ch19L Page 36

_________________________________________

Ch19L Page 37

Problem: An alpha particle and an antiproton are released from rest a great distance apart. They are oppositely charged, so they accelerate towards each other. What are their speeds when they are 2.5 nm apart?

Solution:

Ch19L Page 38

___________________________________________

Problem 37: The electric field has a constant value of 4.0 x 103 V/m 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) 6.0 x 10-3 m directly above P, (b) 3.0 x 10-3 m directly below P, (c) 8.0 x 10-3 m directly to the right of P.

Solution: Remember that the electric field points in the

Ch19L Page 39

Solution: Remember that the electric field points in the direction of DECREASING potential. What do the equipotential surfaces look like for this situation?

(c) Points P and C lie on the same equipotential surface, and so therefore they have the same potential. Thus,

Ch19L Page 40

_____________________________________________

Problem 38: An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate. The plates are separated by a distance of 1.2 cm, and the electric field within the capacitor has a magnitude of 2.1 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Solution: This problem is very similar to Problem 49 from Chapter 18. You might like to compare the solution for this problem to the solution to the other problem in the Chapter 18 lecture notes. They represent two different approaches to very similar problems; decide which approach you like better, once you understand the advantages and disadvantages of each approach.

Here we'll use the principle of conservation of energy. Examine the following diagram, and note that K1 = 0 because the electron is released at rest.

Ch19L Page 41

___________________________________________

This seems like a pretty small number, and it is, but remember that electrons have an extremely tiny mass. To put the electron's kinetic energy in perspective, continue to calculate the electron's speed at position 2.

(Physics majors will note that the speed is about 30% of the speed of light, which means that the entire solution is suspect and should be re-done using relativity. But we'll forgive the textbook for making this error and move on, shall we? We forgive them because our habit is to make things as simple as possible for first-year students, to help them absorb the key concepts, even if not every single detail is exactly correct. But as a physics major, it's good to be alert to such oversimplifications.)

Ch19L Page 42

___________________________________________

Problem: Particle A has positive charge 1.8 nC and is fixed in place. Particle B has positive charge 4.1 nC and is initially placed at rest 2.7 mm from Particle A.(a) Determine the electric potential at the initial position of Particle B.(b) Determine the initial electric potential energy of Particle B.(c) Determine the speed of Particle B when it is far from Particle A. The mass of Particle B is 7.3 x 10-12 kg.

Solution:

Ch19L Page 43

___________________________________________

Problem: Particle A has positive charge 3.2 nC and is fixed in place. Particle B has positive charge 8.3 nC and is also fixed in place. Particle C has positive charge 5.9 nC, is initially located as indicated in the figure, has initial speed 5.3 km/s, and has mass 9.4 x 10-12 kg. (a) Determine the electric potential at the initial position of Particle C.(b) Determine the initial electric potential energy of Particle C.(c) Determine the speed of Particle C when it is far from Particles A and B.

Ch19L Page 44

Solution: Use the superposition principle.

(a) The electric potential at the initial position of Particle C is the sum of the electric potential due to each of Particles A and B. Thus,

(c) Using the principle of conservation of energy,

Ch19L Page 45

Ch19L Page 46