We are used to voltage in our lives— a 12-volt car battery, 110 V or 220 V at home, 1.5 volt flashlight batteries, and so on. Here we see a Van de Graaff generator, whose voltage may reach 50,000 V or more. Voltage is the same as electric potential difference between two points. Electric potential is defined as the potential energy per unit charge.

The children here, whose hair stands on end because each hair has received the same sign of charge, are not harmed by the voltage because the Van de Graaff cannot provide much current before the voltage drops. (It is current through the body that is harmful, as we will see later.)

T £

Electric PotentialCHAPTER-OPENING QUESTIO] —Guess now!Consider a pair of parallel plates with equal and opposite charge densities, a. Which of the following actions will increase the voltage between the plates (assuming fixed charge density)?

(a) Moving the plates closer together.(b) Moving the plates apart.(c) Doubling the area of the plates.(d) Halving the area of the plates.

W e saw in Chapters 7 and 8 that the concept of energy was extremely useful in dealing with the subject of mechanics. The energy point of view is especially useful for electricity. It not only extends the law of conservation of energy, but it gives us another way to view electrical

phenomena. Energy is also a powerful tool for solving Problems more easily in many cases than by using forces and electric fields.

2 3 -1 Electric Potential Energy and Potential Difference

Electric Potential EnergyTo apply conservation of energy, we need to define electric potential energy as we did for other types of potential energy. As we saw in Chapter 8, potential energy can be defined only for a conservative force. The work done by a conservative force in moving an object between any two positions is independent of the path taken. The electrostatic force between anv two charges (Ea. 21-1, F = k 0 1 0o/r2)

CONTENTS23-1 Electric Potential Energy

and Potential Difference 23-2 Relation between Electric

Potential and Electric Field 23-3 Electric Potential Due to

Point Charges 23-4 Potential Due to Any

Charge Distribution 23-5 Equipotential Surfaces 23-6 Electric Dipole Potential 23-7 E Determined from V 23-8 Electrostatic Potential

Energy; the Electron Volt *23-9 Cathode Ray Tube: TV and

Computer Monitors, Oscilloscope

Highpotentialenergy

High ^ potential

Lowpotentialenergy

L̂owpotential

FIGURE 23-1 Work is done by the electric field in moving the positive charge q from position a to position b.

We saw in Chapter 8 that the change in potential energy between two points, a and b, equals the negative of the work done by the conservative force as an object moves from a to b: AU = —W.

Thus we define the change in electric potential energy, Ub — Ua, when a point charge q moves from some point a to another point b, as the negative of the work done by the electric force as the charge moves from a to b. For example, consider the electric field between two equally but oppositely charged parallel plates; we assume their separation is small compared to their width and height, so the field E will be uniform over most of the region, Fig. 23-1. Now consider a tiny positive point charge q placed at point a very near the positive plate as shown. This charge q is so small it has no effect on E. If this charge q at point a is released, the electric force will do work on the charge and accelerate it toward the negative plate. The work W done by the electric field E to move the charge a distance d is

W = Fd = qEdwhere we used Eq. 21-5, F = qE. The change in electric potential energy equals the negative of the work done by the electric force:

Uh - Ua = -w = -q E d [uniform E] (23-1)for this case of uniform electric field E. In the case illustrated, the potential energy decreases (A U is negative); and as the charged particle accelerates from point a to point b in Fig. 23-1, the particle’s kinetic energy K increases—by an equal amount. In accord with the conservation of energy, electric potential energy is transformed into kinetic energy, and the total energy is conserved. Note that the positive charge q has its greatest potential energy at point a, near the positive plate.f The reverse is true for a negative charge: its potential energy is greatest near the negative plate.

ElectricPotential and Potential DifferenceIn Chapter 21, we found it useful to define the electric field as the force per unit charge. Similarly, it is useful to define the electric potential (or simply the potential when “electric” is understood) as the electric potential energy per unit charge. Electric potential is given the symbol V. If a positive test charge q in an electric field has electric potential energy Ua at some point a (relative to some zero poten­tial energy), the electric potential Va at this point is

UaV = — ■a q (23-2a)

As we discussed in Chapter 8, only differences in potential energy are physically meaningful. Hence only the difference in potential, or the potential difference,between two points a and b (such as between a and b in Fig. 23-1) is measurable. When the electric force does positive work on a charge, the kinetic energy increases and the potential energy decreases. The difference in potential energy, Ub — Ua, is equal to the negative of the work, Wha, done by the electric field as the charge moves from a to b; so the potential difference Vha is

Vba = Ay = Fb - Va = Uh ~ U° = - ~ (23-2b)

Note that electric potential, like electric field, does not depend on our test charge q. V depends on the other charges that create the field, not on q; q acquires potential energy by being in the potential V due to the other charges.

We can see from our definition that the positive plate in Fig. 23-1 is at a higher potential than the negative plate. Thus a positively charged object moves naturally from a high potential to a low potential. A negative charge does the reverse.

The unit of electric potential, and of potential difference, is joules/coulomb and is given a special name, the volt, in honor of Alessandro Volta (1745-1827) who is best known for inventing the electric battery. The volt is abbreviated V, so 1 V = 1 J/C. Potential difference, since it is measured in volts, is often referred to as voltage.

If we wish to speak of the potential Va at some point a, we must be aware that Va depends on where the potential is chosen to be zero. The zero for electric potential in a given situation can be chosen arbitrarily, just as for potential energy, because only differences in potential energy can be measured. Often the ground, or a conductor connected directly to the ground (the Earth), is taken as zero potential, and other potentials are given with respect to ground. (Thus, a point where the voltage is 50 V is one where the difference of potential between it and ground is 50 V.) In other cases, as we shall see, we may choose the potential to be zero at an infinite distance (r = oo).

CONCEPTUAL EXAMPLE 25-1 I A negative charge. Suppose a negative charge, such as an electron, is placed near the negative plate in Fig. 23-1, at point b, shown here in Fig. 23-2. If the electron is free to move, will its electric potential energy increase or decrease? How will the electric potential change?RESPONSE An electron released at point b will move toward the positive plate. As the electron moves toward the positive plate, its potential energy decreases as its kinetic energy gets larger, so Ua < Uh and AU = Ua — Uh < 0. But note that the electron moves from point b at low potential to point a at higher potential: Kb = Va — Vb > 0. (Potentials Va and Vh are due to the charges on the plates, not due to the electron.) The sign of AU and AV are opposite because of the negative charge.

Because the electric potential difference is defined as the potential energy difference per unit charge, then the change in potential energy of a charge q when moved between two points a and b is

AU = Ub - [7a = q(Vb - V'a) = c,Vb.A. (23-3)That is, if an object with charge q moves through a potential difference Vba, its potential energy changes by an amount qV\ja. For example, if the potential difference between the two plates in Fig. 23-1 is 6 V, then a +1 C charge moved (say by an external force) from point b to point a will gain (1 C)(6 V) = 6 J of electric potential energy. (And it will lose 6 J of electric potential energy if it moves from a to b.) Similarly, a +2 C charge will gain 12 J, and so on. Thus, electric potential difference is a measure of how much energy an electric charge can acquire in a given situation. And, since energy is the ability to do work, the electric potential difference is also a measure of how much work a given charge can do. The exact amount depends both on the potential difference and on the charge.

To better understand electric potential, let’s make a comparison to the gravitational case when a rock falls from the top of a cliff. The greater the height, h, of a cliff, the more potential energy (= mgh) the rock has at the top of the cliff, relative to the bottom, and the more kinetic energy it will have when it reaches the bottom. The actual amount of kinetic energy it will acquire, and the amount of work it can do, depends both on the height of the cliff and the mass m of the rock. A large rock and a small rock can be at the same height h (Fig. 23-3a) and thus have the same “gravitational potential,” but the larger rock has the greater potential energy (it has more mass). The electrical case is similar (Fig. 23-3b): the potential energy change, or the work that can be done, depends both on the potential difference (corresponding to the height of the cliff) and on the charge (corresponding to mass), Eq. 23-3. But note a significant difference: electric charge comes in two types, + and —, whereas gravitational mass is always +.

Sources of electrical energy such as batteries and electric generators are meant to maintain a potential difference. The actual amount of energy transformed by such a device depends on how much charge flows, as well as the potential difference (Eq. 23-3). For example, consider an automobile headlight connected to a 12.0-V battery. The amount of energy transformed (into light and thermal energy) is proportional to how much charge flows, which depends on how long the light is on. If over a given period of time 5.0 C of charge flows through the light, the total energy transformed is (5.0 C)(12.0 V) = 60J. If the headlight is left on twice as lone. 10.0 C of charge will flow and the energv transformed is

Highpotential

E

Qa b

Lowpotential^ b

High pe for negative charge here

FIGURE 23-2 Central part of Fig. 23-1, showing a negative point charge near the negative plate, where its potential energy ( p e ) is high. Example 23-1.

A CAUTION____________A negative charge has high potential energy when potential V is low

f

V

+ -

V* b̂a v a+ -

h Q+• -•

+2 Q -

+ -

(a) (b)

FIGURE 23-3 (a) Two rocks are at the same height. The larger rock has more potential energy, (b) Two charges have the same electric potential. The 2 Q charge has more potential energy.

TABLE 23-1 Some Typical Potential Differences (Voltages)

SourceVoltage

(approx.)

Thundercloud to ground 108VHigh-voltage power line 105-1 0 6VPower supply for TV tube 104 VAutomobile ignition 104VHousehold outlet 102VAutomobile battery 12 VFlashlight battery 1.5 VResting potential across nerve membrane 10-1 V

Potential changes on skin

h------- vba= --------H5000 V_ +

e~ + +- 0 --- 0)- + +

+ 71/ /

_____ High . ____voltage

FIGURE 23-4 Electron accelerated in CRT. Example 23-2.

Electron in CRT. Suppose an electron in a cathode ray tube (Section 23-9) is accelerated from rest through a potential difference Vb - Va = Vba = +5000 V (Fig. 23-4). (a) What is the change in electric potential energy of the electron? (b) What is the speed of the electron (ra = 9.1 X 10-31 kg) as a result of this acceleration?APPROACH The electron, accelerated toward the positive plate, will decrease in potential energy by an amount A U = qVba (Eq. 23-3). The loss in potential energy will equal its gain in kinetic energy (energy conservation).SOLUTION (a) The charge on an electron is q = - e = - 1.6 X 10-19 C. Therefore its change in potential energy is

AU = qVba = (-1 .6 X 10“19C)(+5000 V) = -8.0 X 10“16J.The minus sign indicates that the potential energy decreases. The potential difference Vba has a positive sign since the final potential Vb is higher than the initial potential Va. Negative electrons are attracted toward a positive electrode and repelled away from a negative electrode.(b) The potential energy lost by the electron becomes kinetic energy K. From conservation of energy (Eq. 8-9a), AK + AU = 0, so

A K = - A U W - 0 = -q(Vb - Fa) = ~qVi,a,

where the initial kinetic energy is zero since we are given that the electron started from rest. We solve for v :

I 2(-1.6 x 10-19 C)(5000 V) ,v = -v / ----------= \ -------------—-----— ---------- = 4.2 X lO'm/s.

M m V 9.1 X 10“31 kgNOTE The electric potential energy does not depend on the mass, only on the charge and voltage. The speed does depend on ra.

EXAMPLE 23-2

FIGURE 23-5 To find Vba in a nonuniform electric field E, we integrate E • d i from point a to point b.

2 3 —2 Relation between Electric Potential and Electric Field

The effects of any charge distribution can be described either in terms of electric field or in terms of electric potential. Electric potential is often easier to use because it is a scalar, as compared to electric field which is a vector. There is a crucial connection between the electric potential produced by a given arrangement of charges and the electric field due to those charges, which we now examine.

We start by recalling the relation between a conservative force F and the potential energy U associated with that force. As discussed in Section 8-2, the difference in potential energy between any two points in space, a and b, is given by Eq. 8-4:

Ub ~ ua F • di,

where di is an infinitesimal increment of displacement, and the integral is taken along any path in space from point a to point b. For the electrical case, we are more interested in the potential difference, given by Eq. 23-2b, Vba = Vb — Va = (Ub — U^/q, rather than in the potential energy itself. Also, the electric field E at any point in space is defined as the force per unit charge (Eq. 21-3): E = F/q. Putting these two relations in the above equation gives us

Vba = Vb - Va = - f E • di. (23-4a)J a

This is the general relation between electric field and potential difference. See Fie. 23-5. If we are siven the electric field due to some arrangement of electric

A simple special case is a uniform field. In Fig. 23-1, for example, a path parallel to the electric field lines from point a at the positive plate to point b at the negative plate gives (since E and dl are in the same direction at each point),

rb rbVba = Vb - Fa = — E • di = - E \ dt = - E d

Ja Jaor

Vba = - E d [only if E is uniform] (23-4b)where d is the distance, parallel to the field lines, between points a and b. Be careful not to use Eq. 23-4b unless you are sure the electric field is uniform.

From either of Eqs. 23-4 we can see that the units for electric field intensity can be written as volts per meter (V/m) as well as newtons per coulomb (N/C). These are equivalent in general, since 1N/C = lN -m /C -m = l J / C - m = lV/m.

EXERCISE A Return to the Chapter-Opening Question, page 607, and answer it again now. Try to explain why you may have answered differently the first time.

EXAMPLE 23-3 Electric field obtained from voltage. Two parallel plates are charged to produce a potential difference of 50 V. If the separation between the plates is 0.050 m, calculate the magnitude of the electric field in the space between the plates (Fig. 23-6).APPROACH We apply Eq. 23-4b to obtain the magnitude of E, assumed uniform. SOLUTION The electric field magnitude is E = Vba/d = (50V/0.050m) = 1000V/m.

EXAMPLE 23-4 Charged conducting sphere. Determine the potential at a distance r from the center of a charged conducting sphere of radius r0 for (a) r > r0, (b) r = r0, (c) r < r0. The total charge on the sphere is Q. APPROACH The charge Q is distributed over the surface of the sphere since it is a conductor. We saw in Example 22-3 that the electric field outside a conducting sphere is

£ = ̂ [r > '■Jand points radially outward (inward if Q < 0). Since we know E, we can start by using Eq. 23-4a.SOLUTION (a) We use Eq. 23-4a and integrate along a radial line with dl parallel to E (Fig. 23-7) between two points which are distances ra and rh from the sphere’s center:

VhL

E • dl = Q

[r*dr

24 ^ 0 Jra r 4ire0

and we set di = dr. If we let V = 0 for r = oo (let’s choose Vh = 0 at rb = oo), then at any other point r (for r > rQ) we have

1 Qv =47T€f > '■o]

We will see in the next Section that this same equation applies for the potential a distance r from a single point charge. Thus the electric potential outside a spherical conductor with a uniformly distributed charge is the same as if all the charge were at its center.(b) As r approaches r0, we see that

F = [r = ro] at the surface of the conductor.(c) For points within the conductor, E = 0. Thus the integral, JE • dl, between r = r0 and any point within the conductor gives zero change in V. Hence V is constant within the conductor:

1 Q

1 b̂a *= 50 V

£ = ?

_d= ^ 5.0 cm

FIGURE 23-6 Example 23-3.

FIGURE 23-7 Example 23-4. Integrating E • dl for the field outside a spherical conductor.

FIGURE 23-8 (a) E versus r, and (b) V versus r, for a positively charged solid conducting sphere of radius r0 (the charge distributes itself on the surface); r is the distance from the center of the sphere.

(a)

FIGURE 23-9 We integrate Eq. 23-4a along the straight line (shown in black) from point a to point b. The line ab is parallel to a field line.

EXAMPLE 23-5 Breakdown voltage. In many kinds of equipment, very high voltages are used. A problem with high voltage is that the air can become ionized due to the high electric fields: free electrons in the air (produced by cosmic rays, for example) can be accelerated by such high fields to speeds sufficient to ionize 0 2 and N2 molecules by collision, knocking out one or more of their electrons. The air then becomes conducting and the high voltage cannot be maintained as charge flows. The breakdown of air occurs for electric fields of about 3 X 106 V/m. (a) Show that the breakdown voltage for a spherical conductor in air is proportional to the radius of the sphere, and (b) estimate the breakdown voltage in air for a sphere of diameter 1.0 cm.APPROACH The electric potential at the surface of a spherical conductor of radius rQ (Example 23-4), and the electric field just outside its surface, are

l Q A l QV = ---------and E = ---------------=-•4776 0 r0 4776 0 n

SOLUTION (a) We combine these two equations and obtainV = r0E. [at surface of spherical conductor]

(b) For r0 = 5 X 10-3 m, the breakdown voltage in air isV = (5 X 10_3m)(3 X 106V/m) » 15,000 V.

When high voltages are present, a glow may be seen around sharp points, known as a corona discharge, due to the high electric fields at these points which ionize air molecules. The light we see is due to electrons jumping down to empty lower states. Lightning rods, with their sharp tips, are intended to ionize the surrounding air when a storm cloud is near, and to provide a conduction path to discharge a dangerous high-voltage cloud slowly, over a period of time. Thus lightning rods, connected to the ground, are intended to draw electric charge off threatening clouds before a large buildup of charge results in a swift destructive lightning bolt.

EXERCISE B On a dry day, a person can become electrically charged by rubbing against rugs and other ordinary objects. Suppose you notice a small shock as you reach for a metal doorknob, noting that the shock occurs along with a tiny spark when your hand is about3.0 mm from the doorknob. As a rough estimate, use Eq. 23-4b to estimate the potential difference between your hand and the doorknob, (a) 9 V; (b) 90 V; (c) 900 V; (d) 9000 V;(e) none of these.

2 3 -3 Electric Potential Due to Point Charges

The electric potential at a distance r from a single point charge Q can be derived directly from Eq. 23-4a, Vh - Va = - f E • dl. The electric field due to a single point charge has magnitude (Eq. 21-4)

1 Q zr / QE = ---------z or E = k —4776 0 r2 r2

(where k = l/47760 = 8.99 X 109N-m2/C 2), and is directed radially outward from a positive charge (inward if Q < 0). We take the integral in Eq. 23-4a along a (straight) field line (Fig. 23-9) from point a, a distance ra from Q, to point b, a distance rb from Q. Then dt will be parallel to E and di = dr. Thus

r b- - q frb i iVa = - E • di = - - d r = —Jr„ 47re0 Jra T2 4w>

Q_Q 47re0 V'b r.

A s m entioned earlier, onlv differences in potential have nhvsical meaning. We are

be whatever we please. It is common to choose the potential to be zero at infinity (let Vh = 0 at rb = oo). Then the electric potential V at a distance r from a single point charge is

T/ = 1 Q [single point charge;]V 47reo r [ V = 0 at r = oo J

We can think of V here as representing the absolute potential, where V = 0 at r = oo, or we can think of V as the potential difference between r and infinity. Notice that the potential V decreases with the first power of the distance, whereas the electric field (Eq. 21-4) decreases as the square of the distance. The potential near a positive charge is large, and it decreases toward zero at very large distances (Fig. 23-10). For a negative charge, the potential is negative and increases toward zero at large distances (Fig. 23-11).

In Example 23-4 we found that the potential due to a uniformly charged sphere is given by the same relation, Eq. 23-5, for points outside the sphere. Thus we see that the potential outside a uniformly charged sphere is the same as if all the charge were concentrated at its center.

I EXERCISE C What is the potential at a distance of 3.0 cm from a point charge Q = -2 .0 X 10“9C? (a) 600 V; (b) 60 V; (c) 6 V; (d) -6 0 0 V; (e) -6 0 V; ( / ) - 6 V.

Work required to bring two positive charges close together.What minimum work must be done by an external force to bring a charge q = 3.00 fiC from a great distance away (take r = oo ) to a point 0.500 m from a charge Q = 20.0 /jlC?

APPROACH To find the work we cannot simply multiply the force times distance because the force is not constant. Instead we can set the change in potential energy equal to the (positive of the) work required of an external force (Chapter 8), and Eq. 23-3: W = AC/ = q(Vb - Va). We get the potentials Vh and Va using Eq. 23-5.

SOLUTION The work required is equal to the change in potential energy:

EXAMPLE 23-6

w = q ( v b - Va)

kQ kQr\> ra

where rh = 0.500 m and ra = oo. The right-hand term within the parentheses iszero ( l /o o = 0) so

, N (8.99 X 109N -m 7C 2)(2.00 X 10“5C)W = (3.00 X 10 C ------------------^ ^ 7;------------------ = 1.08 J.v ' (0.500 m)

NOTE We could not use Eq. 23-4b here because it applies only to uniform fields.But we did use Eq. 23-3 because it is always valid.

To determine the electric field at points near a collection of two or more point charges requires adding up the electric fields due to each charge. Since the electric field is a vector, this can be time consuming or complicated. To find the electric potential at a point due to a collection of point charges is far easier, since the electric potential is a scalar, and hence you only need to add numbers (with appropriate signs) without concern for direction. This is a major advantage in using electric ootential for solving Problems.

FIGURE 23-10 Potential V as a function of distance r from a single point charge Q when the charge is positive.

FIGURE 23-11 Potential V as a function of distance r from a single point charge Q when the charge is negative.

/?\ CAUTIONWe cannot use W = Fd when F is not constant

FIGURE 23-12 Example 23-7. (See also Example 21-8, Fig. 21-27.)

CAUTIONPotential is a scalar and

has no components

© <

(i)

© Q

(ii)

® ©

(iii)

FIGURE 23-13 Exercise D.

EXAMPLE 23-7 Potential above two charges. Calculate the electric potential(a) at point A in Fig. 23-12 due to the two charges shown, and (b) at point B. [This is the same situation as Example 21-8, Fig. 21-27, where we calculated the electric field at these points.]APPROACH The total potential at point A (or at point B) is the sum of the potentials at that point due to each of the two charges Q1 and Q2. The potential due to each single charge is given by Eq. 23-5. We do not have to worry about directions because electric potential is a scalar quantity. But we do have to keep track of the signs of charges.SOLUTION (a) We add the potentials at point A due to each charge Q1 and Q2, and we use Eq. 23-5 for each:

VA = VA2 + l^i, Q2 , , Qi= k ------- 1- k —

r2A r1Awhere r1A = 60 cm and r2A = 30 cm. Then

VA =(9.0 X 109N-m2/C 2)(5.0 X 10“5C)

0.30 m(9.0 X 109 N • m2/C 2)(—5.0 X 10“5C)

(b) At point B,

0.60 m1.50 X 106V - 0.75 X 106V7.5 X 105V.

0.40 m, sorlB ~ r2B

= VB2 + VpB1

(9.0 X 109N-m2/C 2)(5.0 X 10“5C)0.40 m

(9.0 X 109 N • m2/C 2)( -5.0 X 10“5C)0.40 m

= 0V.NOTE The two terms in the sum in (b) cancel for any point equidistant from Qi and Q2 (r1B = r2B). Thus the potential will be zero everywhere on the plane equidistant between the two opposite charges. This plane where V is constant is called an equipotential surface.

Simple summations like these can easily be performed for any number of point charges.

EXERCISE D Consider the three pairs of charges, Q\ and Q2, in Fig. 23-13. (a) Which set has a positive potential energy? (b) Which set has the most negative potential energy?(c) Which set requires the most work to separate the charges to infinity? Assume the charges all have the same magnitude.

2 3 -4 Potential Due to Any Charge Distribution

If we know the electric field in a region of space due to any distribution of electric charge, we can determine the difference in potential between two points in the region using Eq. 23-4a, Vba = - Ja E • dt. In many cases we don’t know E as a function of position, and it may be difficult to calculate. We can calculate the potential V due to a given charge distribution in another way, using the potential due to a single point charge, Eq. 23-5:

l Qv =47T€f

If we have n individual point charges, the potential at some point a (relative toV = 0 at r = oo) is

n 1 n q .

^ . = 2 ^ = 4 (23“6a>j=1 4776 o i=i

where ria is the distance from the ith charge (Qt) to the point a. (We already used this approach in Example 23-7.) If the charge distribution can be considered continuous, then

V = — f — > (23-6b)4 i r e 0 J r

where r is the distance from a tiny element of charge, dq, to the point where V is being determined.

IfrfiVilJ Potential due to a ring of charge. A thin circular ring ofradius R has a uniformly distributed charge Q. Determine the electric potential at a point P on the axis of the ring a distance x from its center, Fig. 23-14.

APPROACH We integrate over the ring using Eq. 23-6b.SOLUTION Each point on the ring is equidistant from point P, and this distance is (x2 + R2)2. So the potential at P is:

V =1 (d q = 1 1 f

4 7 T € 0 J r 4 7 7 6 o (x2 + R2)i J q 4 7 7 6 o (x2 + R2)\

NOTE For points very far away from the ring, x » R, this result reduces to (l/477€0) (£?/*), the potential of a point charge, as we should expect.

EXAMPLE 23-9 Potential due to a charged disk. A thin flat disk, of radius R0, has a uniformly distributed charge Q, Fig. 23-15. Determine the potential at a point P on the axis of the disk, a distance x from its center.

APPROACH We divide the disk into thin rings of radius R and thickness dR and use the result of Example 23-8 to sum over the disk.SOLUTION The charge Q is distributed uniformly, so the charge contained in each ring is proportional to its area. The disk has area ttRq and each thin ring has area dA = (2irR)(dR). Hence

dq 2ttR dRQ nRl

so(2irR){dR) 2 QRdR

dq = Q

Then the potential at P, using Eq. 23-6b in which r is replaced by (x2 + R2)2, is

l _ f dq = 2 Q f*° R dR = Q 1

T € 0 J ( X 2 + R 2)l 4 7 7 € 0 i ? o JO (X2 + R 2) 2 2 7 7 € 0 i ? o

R=Ro

2776 0 R q

NOTE For x i?0; this formula reduces to

[*277€0 i?0 L V ^ X21 + i ^ \ - x\ =

FIGURE 23-14 Example 23-8. Calculating the potential at point P, a distance x from the center of a uniform ring of charge.

FIGURE 23-15 Example 23-9. Calculating the electric potential at point P on the axis of a uniformly charged thin disk.

2 3 —5 Equipotential Surfaces

20V.

FIGURE 23-16 Equipotential lines (the green dashed lines) between two oppositely charged parallel plates. Note that they are perpendicular to the electric field lines (solid red lines).

The electric potential can be represented graphically by drawing equipotential lines or, in three dimensions, equipotential surfaces. An equipotential surface is one on which all points are at the same potential. That is, the potential difference between any two points on the surface is zero, and no work is required to move a charge from one point to the other. An equipotential surface must be perpendicular to the electric field at any point. If this were not so—that is, if there were a component of E parallel to the surface—it would require work to move the charge along the surface against this component of E; and this would contradict the idea that it is an equipotential surface. This can also be seen from Eq. 23-4a, AV = - JE • dl. On a surface where V is constant, AV = 0, so we must have either E = 0, di = 0, or cos 0 = 0 where 0 is the angle between E and di. Thus in a region where E is not zero, the path di along an equipotential must have cos0 = 0, meaning 6 = 90° and E is perpendicular to the equipotential.

The fact that the electric field lines and equipotential surfaces are mutually perpendicular helps us locate the equipotentials when the electric field lines are known. In a normal two-dimensional drawing, we show equipotential lines, which are the intersections of equipotential surfaces with the plane of the drawing. In Fig. 23-16, a few of the equipotential lines are drawn (dashed green lines) for the electric field (red lines) between two parallel plates at a potential difference of 20 V. The negative plate is arbitrarily chosen to be zero volts and the potential of each equipotential line is indicated. Note that E points toward lower values of V.

FIGURE 23-17 Example 23-10. Electric field lines and equipotential surfaces for a point charge.

EXAMPLE 23-10 Point charge equipotential surfaces. For a single point charge with Q = 4.0 X 10-9 C, sketch the equipotential surfaces (or lines in a plane containing the charge) corresponding to Vx = 10 V, V2 = 20 V, and y3 = 30 V.

APPROACH The electric potential V depends on the distance r from the charge (Eq. 23-5).SOLUTION The electric field for a positive point charge is directed radially outward. Since the equipotential surfaces must be perpendicular to the lines of electric field, they will be spherical in shape, centered on the point charge, Fig. 23-17. From Eq. 23-5 we have r = (1/47re0)(Q/V), so that for Vi = 10 V, rx = (9.0 X 109 N • m2/C 2)(4.0 X Hn9C)/(10V) = 3.6 m, for V2 = 20 V, r2 = 1.8 m, and for V3 = 30 V, r3 = 1.2 m, as shown.NOTE The equipotential surface with the largest potential is closest to the positive charge. How would this change if Q were negative?

FIGURE 23-18 Equipotential lines (green, dashed) are always perpendicular to the electric field lines (solid red) shown here for two equal but oppositely charged particles.

The equipotential lines for the case of two equal but oppositely charged particles are shown in Fig. 23-18 as green dashed lines. Equipotential lines and surfaces, unlike field lines, are always continuous and never end, and so continue beyond the borders of Figs. 23-16 and 23-18.

We saw in Section 21-9 that there can be no electric field within a conductor in the static case, for otherwise the free electrons would feel a force and would move. Indeed, the entire volume of a conductor must be entirely at the same potential in the static case, and the surface of a conductor is then an equipotential surface. (If it weren’t, the free electrons at the surface would move, since whenever there is a potential difference between two points, free charges will move.) This is fully consistent with our result, discussed earlier, that the electric field at the surface of a conductor must be perpendicular to

A useful analogy for equipotential lines is a topographic map: the contour lines are essentially gravitational equipotential lines (Fig. 23-19).

JStit'ilr U ik t

FIGURE 23-19 A topographic map (here, a portion of the Sierra Nevada in California) shows continuous contour lines, each of which is at a fixed height above sea level. Here they are at 80 ft (25 m) intervals. If you walk along one contour line, you neither climb nor descend. If you cross lines, and especially if you climb perpendicular to the lines, you will be changing your gravitational potential (rapidly, if the lines are close together).

23—6 Electric Dipole PotentialTwo equal point charges Q, of opposite sign, separated by a distance i, are called an electric dipole, as we saw in Section 21-11. Also, the two charges we saw in Figs. 23-12 and 23-18 constitute an electric dipole, and the latter shows the electric field lines and equipotential surfaces for a dipole. Because electric dipoles occur often in physics, as well as in other fields, it is useful to examine them more closely.

The electric potential at an arbitrary point P due to a dipole, Fig. 23-20, is the sum of the potentials due to each of the two charges (we take V = 0 at r = oo):

1 0 , 1 (-Q) 1 Jl 1 \ Q A rV = 47reo r 47T€0 (r + Ar) 47T€0 ® \r r + Ar 47T€0 r(r + Ar)where r is the distance from P to the positive charge and r + Ar is the distance to the negative charge. This equation becomes simpler if we consider points P whose distance from the dipole is much larger than the separation of the two charges— that is, for r » t From Fig. 23-20 we see that Ar « i cos 0; since r » Ar = i cos 0, we can neglect Ar in the denominator as compared to r. Therefore, we obtain

1 Qi cos 6 1 p cos 0V = - ---- [dipole; r » I] (23-7)

47T€o r 4776 o rwhere p = Qi is called the dipole moment. We see that the potential decreases as the square of the distance from the dipole, whereas for a single point charge the potential decreases with the first power of the distance (Eq. 23-5). It is not surprising that the potential should fall off faster for a dipole; for when you are far from a dipole, the two equal but opposite charges appear so close together as to tend to neutralize each other.

Table 23-2 gives the dipole moments for several molecules. The + and - signs indicate on which atoms these charges lie. The last two entries are a part of many organic molecules and play an important role in molecular biology. A dipole moment has units of coulomb-meters (C-m), although for molecules a smaller unit called a debye is sometimes used: 1 debye = 3.33 X IO-30 C • m.

23—7 E Determined from VWe can use Eq. 23-4a, Vb — Va = — E • di, to determine the difference in potential between two points if the electric field is known in the region between those two points. By inverting Eq. 23-4a, we can write the electric field in terms of the Dotential. Then the electric field can be determined from a knowledge of V.

FIGURE 23-20 Electric dipole. Calculation of potential V at point P.

TABLE 23-2 Dipole Moments of Selected Molecules

MoleculeD ipole Moment

(C m)

H2(+)o (“) 6.1 X 10'-30

H<+>C1<“> 3.4 X 10'-30

N(-)h 3(+) 5.0 X 10“-30

>N<“> — H<+> W3.01, X 10“-30

> C (+) = Q(-) W8.01, X 10“-30

1 These groups often appear on larger molecules; hence the value for the dipole moment will vary somewhat, depending on the rest of the molecule.

We write Eq. 23-4a in differential form as dV = -E • dl = -E ,d i ,

where dV is the infinitesimal difference in potential between two points a distance di apart, and Et is the component of the electric field in the direction of the infinitesimal displacement di. We can then write

dV£ , = - - ■ (23-8)

Thus the component o f the electric field in any direction is equal to the negative o f the rate o f change o f the electric potential with distance in that direction. The quantity dV/di is called the gradient of V in a particular direction. If the direction is not specified, the term gradient refers to that direction in which V changes most rapidly; this would be the direction of E at that point, so we can write

E = ~ ~ [if d?||E]

If E is written as a function of x, y, and z, and we let i refer to the x, y, and z axes, then Eq. 23-8 becomes

dV dV dVEx = ------ > Ev = -------> Ez = -------- (23-9)* dx y dy z dz

Here, dV/ dx is the “partial derivative” of V with respect to x, with y and z held constant.1For example, if V (x ,y ,z ) = (2V/m2)x2 + (8V/m3)^2z + (2V/m2)z2, then

Ex = -dV /dx = - (4 V /m 2)x,Ey = -dV /dy = —(16 V/m 3)yz,

andEz = -dV /dz = - (8 V /m 3)y2 - (4V/m2)z.

U 5 E E I H H H E for ring and disk. Use electric potential to determine the electric field at point P on the axis of (a) a circular ring of charge (Fig. 23-14) and (b) a uniformly charged disk (Fig. 23-15).APPROACH We obtained V as a function of x in Examples 23-8 and 23-9, so we find E by taking derivatives (Eqs. 23-9).SOLUTION (a) From Example 23-8,

V = Q4^6 o (jc2 + R2)i

Then_ d V = 1 Qx

dx 47re0 {x2 + R2)l

This is the same result we obtained in Example 21-9.(b) From Example 23-9,

v - +so

dV Qdx 2ire 0Rq [ (x2 +

For points very close to the disk, x « R0, this can be approximated by

E « 6 =2ire0i?o 2e0

where a = Q /ttRq is the surface charge density. We also obtained these results in Chapter 21, Example 21-12 and Eq. 21-7.

Equation 23-9 can be written as a vector equation,

If we compare this last Example with Examples 21-9 and 21-12, we see that here, as for many charge distributions, it is easier to calculate V first, and then E from Eq. 23-9, rather than to calculate E due to each charge from Coulomb’s law. This is because V due to many charges is a scalar sum, whereas E is a vector sum.

2 3 —8 Electrostatic Potential Energy; the Electron Volt

Suppose a point charge q is moved between two points in space, a and b, where the electric potential due to other charges is Va and Vb, respectively. The change in electro­static potential energy of q in the field of these other charges is, according to Eq. 23-2b,

A U = Ub - U , = q{Vb - V̂ ).Now suppose we have a system of several point charges. What is the electrostatic

potential energy of the system? It is most convenient to choose the electric potential energy to be zero when the charges are very far (ideally infinitely far) apart. A single point charge, Q1, in isolation, has no potential energy, because if there are no other charges around, no electric force can be exerted on it. If a second point charge Q2 is brought close to Q1, the potential due to Q1 at the position of this second charge is

4ire0 rnwhere r12 is the distance between the two. The potential energy of the two charges, relative to V = 0 at r = oo, is

u = Q2V = ------(23-10)4 ire0 r12This represents the work that needs to be done by an external force to bring Q2 from infinity (V = 0) to a distance r12 from Qx. It is also the negative of the work needed to separate them to infinity.

If the system consists of three charges, the total potential energy will be the work needed to bring all three together. Equation 23-10 represents the work needed to bring Q2 close to Qx; to bring a third charge Q3 so that it is a distance r13 from Qx and r23 from Q2 requires work equal to

1 Q 1 Q 3 1 Q 2 Q 3

4/7T€0 r13 4/7T€0 r23 So the potential energy of a system of three point charges is

T J 1 ( Q l Q 2 Q l Q 3 Q l Q 3 \ r-r j n nU = ------------------1----------- 1--------- I. \V = 0 at r = 00147760 V r 1 2 r13 r23 J

For a system of four charges, the potential energy would contain six such terms, and so on. (Caution must be used when making such sums to avoid double counting of the different pairs.)

The Electron Volt UnitThe joule is a very large unit for dealing with energies of electrons, atoms, or molecules (see Example 23-2), and for this purpose, the unit electron volt (eV) is used. One electron volt is defined as the energy acquired by a particle carrying a charge whose magnitude equals that on the electron (q = e) as a result of moving through a potential difference of 1V. Since e = 1.6 X 10-19 C, and since the change in potential energy equals qV, 1 eV is equal to (1.6 X 10-19 C)(1.0 V) = 1.6 X 10-19 J:

leV = 1.6 X 10_19J.An electron that accelerates through a potential difference of 1000 V will lose 1000 eV of potential energy and will thus gain 1000 eV or 1 keV (kiloelectron volt) of kinetic energy. On the other hand, if a particle with a charge equal to twice the magnitude of the charge on the electron (= 2e = 3.2 X 10-19C) moves through a

FIGURE 23-21 If the cathode inside the evacuated glass tube is heated to glowing, negatively charged “cathode rays” (electrons) are “boiled off” and flow across to the anode (+ ) to which they are attracted.

Cathode Anode

nr

Although the electron volt is handy for stating the energies of molecules and elementary particles, it is not a proper SI unit. For calculations it should be converted to joules using the conversion factor given above. In Example 23-2, for example, the electron acquired a kinetic energy of 8.0 X 10-16J. We normally would quote this energy as 5000 eV (= 8.0 X 10_16J/1.6 X 10_19J/eV). But when determining the speed of a particle in SI units, we must use the kinetic energy in J.

I EXERCISE E What is the kinetic energy of a He2+ ion released from rest and accelerated through a potential difference of 1.0 kV? {a) 1000 eV, (b) 500 eV, (c) 2000 eV, (d) 4000 eV, (e) 250 eV.

Disassembling a hydrogen atom. Calculate the work needed to “disassemble” a hydrogen atom. Assume that the proton and electron are initially separated by a distance equal to the “average” radius of the hydrogen atom in its ground state, 0.529 X 10 10 m, and that they end up an infinite distance apart from each other. APPROACH The work necessary will be equal to the total energy, kinetic plus potential, of the electron and proton as an atom, compared to their total energy when infinitely far apart.SOLUTION From Eq. 23-10 we have initially

1 Q\Qi _ 1 (e)(-« ) _ ~ (8-99 x 109N • m2/C 2)(l.60 X 1(T19C)2 u 47T£0 r 4ire„ r (0.529 X 10“10m)

= -27.2(1.60 X 10“19) J = -27.2 eV.This represents the potential energy. The total energy must include also the kinetic energy of the electron moving in an orbit of radius r = 0.529 X 10_10m. From F = ma for centripetal acceleration, we have

1 / e2\ _ mv2 47760 \ r 2J r

ThenK = \m v2 = \

which equals - \U (as calculated above), so K = +13.6eV. The total energy initially is E = K + U = 13.6 eV - 27.2 eV = -13.6 eV. To separate a stable hydrogen atom into a proton and an electron at rest very far apart (U = 0 at r = oo, K = 0 because v = 0) requires +13.6 eV. This is, in fact, the measured ionization energy for hydrogen. NOTE To treat atoms properly, we need to use quantum theory (Chapters 37 to 39). But our “classical” calculation does give the correct answer here.

I EXERCISE F The kinetic energy of a 1000-kg automobile traveling 20 m /s (70 km/h) | would be about (a) 100 GeV, (b) 1000 TeV, (c) 106TeV, (d) 1012TeV, (e) 1018TeV.

* 2 3 -9 Cathode Ray Tube: IV and Computer Monitors, Oscilloscope

An important device that makes use of voltage, and that allows us to “visualize” how a voltage changes in time, is the cathode ray tube (CRT). A CRT used in this way is an oscilloscope. The CRT has also been used for many years as the picture tube of television sets and computer monitors, but LCD (Chapter 35) and other screens are now common.

The operation of a CRT depends on the phenomenon of thermionic emission discovered by Thomas Edison (1847-1931). Consider two small plates (electrodes) inside an evacuated “bulb” or “tube” as shown in Fig. 23-21, to which is applied a potential difference. The negative electrode is called the cathode, the positive one the anode. If the negative cathode is heated (usually by an electric current, as in a lightbulb) so that it becomes hot and glowing, it is found that negative charge leaves the cathode and flows to the positive anode. These negative charges are now called electrons, but originally they were called cathode rays since

EXAMPLE 23-12

deflectionplates

FIGURE 23-22 A cathode ray tube. Magnetic deflection coils are often used in place of the electric deflection plates shown here. The relative positions of the elements have been exaggerated for clarity.

The cathode ray tube (CRT) derives its name from the fact that inside an evacuated glass tube, a beam of cathode rays (electrons) is directed to various parts of a screen to produce a “picture.” A simple CRT is diagrammed in Fig. 23-22. Electrons emitted by the heated cathode are accelerated by a high voltage (5000-50,000 V) applied between the anode and cathode. The electrons pass out of this “electron gun” through a small hole in the anode. The inside of the tube face is coated with a fluorescent material that glows when struck by electrons. A tiny bright spot is thus visible where the electron beam strikes the screen. Two horizontal and two vertical plates can deflect the beam of electrons when a voltage is applied to them. The electrons are deflected toward whichever plate is positive. By varying the voltage on the deflection plates, the bright spot can be placed at any point on the screen. Many CRTs use magnetic deflection coils (Chapter 27) instead of electric plates.

In the picture tube or monitor for a computer or television set, the electron beam is made to sweep over the screen in the manner shown in Fig. 23-23 by changing voltages applied to the deflection plates. For standard television in the United States, 525 lines constitutes a complete sweep in ^ s, over the entire screen. High-definition TV provides more than double this number of lines (1080), giving greater picture sharpness. We see a picture because the image is retained by the fluorescent screen and by our eyes for about ^ s. The picture we see consists of the varied brightness of the spots on the screen, controlled by the grid (a “porous” electrode, such as a wire grid, that allows passage of electrons). The grid limits the flow of electrons by means of the voltage (the “video signal”) applied to it: the more negative this voltage, the more electrons are repelled and the fewer pass through. This video signal sent out by the TV station, and received by the TV set, is accompanied by signals that synchronize the grid voltage to the horizontal and vertical sweeps. (More in Chapter 31.)

An oscilloscope is a device for amplifying, measuring, and visually observing an electrical signal as a function of time on the screen of a CRT (a “signal” is usually a time-varying voltage). The electron beam is swept horizontally at a uniform rate in time by the horizontal deflection plates. The signal to be displayed is applied (after amplification) to the vertical deflection plates. The visible “trace” on the screen, which could be an electrocardiogram (Fig. 23-24), or a signal from an experiment on nerve conduction, is a plot of the signal voltage (vertically) versus time (horizontally).

( ^ P H Y S I C S A P P L I E DCRT, TV and computer monitors

FIGURE 23-23 Electron beam sweeps across a television screen in a succession of horizontal lines.Each horizontal sweep is made by varying the voltage on the horizontal deflection plates. Then the electron beam is moved down a short distance by a change in voltage on the vertical deflection plates, and the process is repeated.

P H Y S I C S A P P L I E DOscilloscope

FIGURE 23-24 Anelectrocardiogram (ECG) trace displayed on a CRT.

SummaryElectric potential is defined as electric potential energy per unit charge. That is, the electric potential difference between any two points in space is defined as the difference in potential energy of a test charge q placed at those two points, divided by the charge q:

Uh - UaVba = (23-2b)

Potential difference is measured in volts (IV = 1J/C) and is sometimes referred to as voltage.

The change in potential energy of a charge q when it moves through a potential difference Vj,a is

A U = qVha. (23-3)

The potential difference Vba between two points, a and b, is given by the relation

Vba = Vb - Va = - j E - d l (23-4a)

Thus Vba can he found in any region where E is known. If the electric field is uniform, the integral is easy: Vba = —Ed,

where d is the distance (parallel to the field lines) between the two points.

An equipotential line or surface is all at the same potential, and is perpendicular to the electric field at all points.

The electric potential due to a single point charge Q, relative to zero potential at infinity, is given by

V = - 1— - • (23-5)4-7760 rThe potential due to any charge distribution can be obtained by summing (or integrating) over the potentials for all the charges.

The potential due to an electric dipole drops off as 1/r2. The dipole moment is p = Qi, where i is the distance between the two equal but opposite charges of magnitude Q.

When V is known, the components of E can be found from the inverse of Eq. 23-4a, namely

Ex = - ^ 7 ’ Ey = Ez = (23-9)dx y dy * dz[*Television and computer monitors traditionally use a cathode ray tube (CRT) that accelerates electrons by high voltage, and sweeps them across the screen in a regular way using deflection plates.]

Questions1. If two points are at the same potential, does this mean that no

work is done in moving a test charge from one point to the other? Does this imply that no force must be exerted? Explain.

2. If a negative charge is initially at rest in an electric field, will it move toward a region of higher potential or lower poten­tial? What about a positive charge? How does the potential energy of the charge change in each instance?

3. State clearly the difference (a) between electric potential and electric field, (b) between electric potential and electric potential energy.

4. An electron is accelerated by a potential difference of, say,0.10 V. How much greater would its final speed be if it is accelerated with four times as much voltage? Explain.

5. Can a particle ever move from a region of low electric potential to one of high potential and yet have its electric potential energy decrease? Explain.

6. If V = 0 at a point in space, must E = 0? If E = 0 at some point, must V = 0 at that point? Explain. Give exam­ples for each.

7. When dealing with practical devices, we often take the ground (the Earth) to be 0 V. (a) If instead we said the ground was —10 V, how would this affect V and E at other points? (b) Does the fact that the Earth carries a net charge affect the choice of V at its surface?

8. Can two equipotential lines cross? Explain.

9. Draw in a few equipotential lines in Fig. 21-34b and c.

10. What can you say about the electric field in a region of space that has the same potential throughout?

11. A satellite orbits the Earth alona a gravitational enuinoten-

12. Suppose the charged ring of Example 23-8 was not uniformly charged, so that the density of charge was twice as great near the top as near the bottom. Assuming the total charge Q is unchanged, would this affect the potential at point P on the axis (Fig. 23-14)? Would it affect the value of E at that point? Is there a discrepancy here? Explain.

13. Consider a metal conductor in the shape of a football. If it carries a total charge Q, where would you expect the charge density cr to be greatest, at the ends or along the flatter sides? Explain. [Hint. Near the surface of a conductor, E = o-/e0.]

14. If you know V at a point in space, can you calculate E at that point? If you know E at a point can you calculate V at that point? If not, what else must be known in each case?

15. A conducting sphere carries a charge Q and a second iden­tical conducting sphere is neutral. The two are initially isolated, but then they are placed in contact, (a) What can you say about the potential of each when they are in contact? (b) Will charge flow from one to the other? If so, how much? (c) If the spheres do not have the same radius, how are your answers to parts (a) and (b) altered?

16. At a particular location, the electric field points due north. In what direction(s) will the rate of change of potential be(a) greatest, (b) least, and (c) zero?

17. Equipotential lines are spaced 1.00 V apart. Does the distance between the lines in different regions of space tell you anything about the relative strengths of E in those regions? If so, what?

18. If the electric field E is uniform in a region, what can you infer about the electric potential V? If V is uniform in a region of space, what can you infer about E?

19. Is the electric potential energy of two unlike charges posi­tive or negative? W hat about two like charges? W hat is the

| Problems2 3 -1 Electric Potential1. (I) What potential difference is needed to stop an electron

that has an initial velocity v = 5.0 X 105 m/s?2. (I) How much work does the electric field do in moving a

proton from a point with a potential of +185 V to a point where it is -55 V?

3. (I) An electron acquires 5.25 X IO-16 J of kinetic energy when it is accelerated by an electric field from plate A to plate B. What is the potential difference between the plates, and which plate is at the higher potential?

4. (II) The work done by an external force to move a -9 .10 /aC charge from point a to point b is 7.00 X 10-4 J. If the charge was started from rest and had 2.10 X 10_4J of kinetic energy when it reached point b, what must be the potential difference between a and b?

2 3 -2 Potential Related to Electric Field5. (I) Thunderclouds typically develop voltage differences of

about 1 X 108V. Given that an electric field of 3 X 106V/m is required to produce an electrical spark within a volume of air, estimate the length of a thunder­cloud lightning bolt. [Can you see why, when lightning strikes from a cloud to the ground, the bolt actually has to propagate as a sequence of steps?]

6. (I) The electric field between two parallel plates connected to a 45-V battery is 1300 V/m. How far apart are the plates?

7. (I) What is the maximum amount of charge that a spherical conductor of radius 6.5 cm can hold in air?

8. (I) What is the magnitude of the electric field between two parallel plates 4.0 mm apart if the potential difference between them is 110 V?

9. (I) What minimum radius must a large conducting sphere of an electrostatic generating machine have if it is to be at35,000 V without discharge into the air? How much charge will it carry?

10. (II) A manufacturer claims that a carpet will not generate more than 5.0 kV of static electricity. What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a 5.0-kV potential difference between the shoe and the carpet, approximating the shoe and the carpet as large sheets of charge separated by a distance d = 1.0 mm?

11. (II) A uniform electric field E = —■4.20 N/Ci points in the negative x direction as shown in Fig. 23-25. The x and y coordinates of points A, B, and C are given on the diagram (in meters). Determinethe differences in ^potential (a) VBA, C(-3.00 ,4.00)(b)V{cb, and (c)yCA- --------<----

B(4.00, 4.00)

A(4.00, 1.00) ------^x

FIGURE 23-25 *Problem 11.

12. (II) The electric potential of a very large isolated flat metal plate is Vq. It carries a uniform distribution of charge of surface density a (C/m2), or o-/2 on each surface. Determine V at a distance x from the plate. Consider thenoint r to he far from the erlaes and assume r is mnr.h

13. (II) The Earth produces an inwardly directed electric field of magnitude 150 V/m near its surface, (a) What is the potential of the Earth’s surface relative to V = 0 at r = oo? (b) If the potential of the Earth is chosen to be zero, what is the potential at infinity? (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth’s net charge; how would this affect your answer?)

14. (II) A 32-cm-diameter conducting sphere is charged to 680 V relative to V = 0 at r = oo. (a) What is the surface charge density crl (b) At what distance will the potential due to the sphere be only 25 V?

15. (II) An insulated spherical conductor of radius rx carries a charge Q. A second conducting sphere of radius r2 and initially uncharged is then connected to the first by a long conducting wire, (a) After the connection, what can you say about the electric potential of each sphere? (b) How much charge is transferred to the second sphere? Assume the connected spheres are far apart compared to their radii. (Why make this assumption?)

16. (II) Determine the difference in potential between two points that are distances Ra and Rb from a very long ( » Raor Rb) straight wire carrying a uniform charge per unit length A.

17. (II) Suppose the end of your finger is charged, (a) Estimate the breakdown voltage in air for your finger. (b) About what surface charge density would have to be on your finger at this voltage?

18. (II) Estimate the electric field in the membrane wall of a living cell. Assume the wall is 10 nm thick and has a poten­tial of 0.10 V across it.

19. (II) A nonconducting sphere of radius r0 carries a total charge Q distributed uniformly throughout its volume. Determine the electric potential as a function of the distance r from the center of the sphere for (a) r > r0 and (b) r < r0. Take V = 0 at r = oo. (c) Plot V versus r and E versus r.

20. (Ill) Repeat Problem 19 assuming the charge density pE increases as the square of the distance from the center of the sphere, and pE = 0 at the center.

21. (Ill) The volume charge density pE within a sphere of radius rQ is distributed in accordance with the following spherically symmetric relation

PE(r) = Po[l -

where r is measured from the center of the sphere and p0 is a constant. For a point P inside the sphere (r < r0), determine the electric potential V. Let V = 0 at infinity.

22. (Ill) A hollow spherical conductor, carrying a net charge +Q, has inner radius r\ and outer radius r2 = 2 (Fig. 23-26). At the center of the sphere is a point charge +<2/2.(a) Write the electric field strength E in all three regions as a function of r. Then determine the potential as a function of r, the distance from the center, for (b) r > r2, (c) ri < r < r2, and(d) 0 < r < r1. (e) Plot both V and E as a function of r from r = 0 to r = 2r2.

FIGURE 23-26

23. (Ill) A very long conducting cylinder (length £) of radius R0 (Rq « t) carries a uniform surface charge density cr (C/m2). The cylinder is at an electric potential VQ. What is the poten­tial, at points far from the end, at a distance R from the center of the cylinder? Determine for (a) R > R0 and(b) R < R0. (c) Is V = 0 at R = oo (assume £ = oo)? Explain.

23-3 Potential Due to Point Charges24. (I) A point charge Q creates an electric potential of +185 V

at a distance of 15 cm. What is Q (let V = 0 at r = oo) ?25. (I) (a) What is the electric potential 0.50 X 10-10 m from a

proton (charge +e)l Let V = 0 at r = oo. (b) What is the potential energy of an electron at this point?

26. (II) Two point charges, 3.4/iC and — 2.0/iC, are placed5.0 cm apart on the x axis. At what points along the x axis is(a) the electric field zero and (b) the potential zero? LetV = 0 at r = oo.

27. (II) A +25 /jlC point charge is placed 6.0 cm from an iden­tical +25 /jlC point charge. How much work would be required by an external force to move a +0.18 /iC test charge from a point midway between them to a point 1.0 cm closer to either of the charges?

28. (II) Point a is 26 cm north of a -3.8 /jlC point charge, and point b is 36 cm west of the charge (Fig. 23-27). Determine(a) Vi, - Va, and (b) Eb - Ea(magnitude and direction).

26 cmFIGURE 23-27 b__Problem 28. 36 cm Q = -3.8 /iC

29. (II) How much voltage must be used to accelerate a proton (radius 1.2 X 10-15 m) so that it has sufficient energy to just “touch” a silicon nucleus? A silicon nucleus has a charge of +14e and its radius is about 3.6 X 10-15m. Assume the potential is that for point charges.

30. (II) Two identical +5.5 /jlC point charges are initially spaced6.5 cm from each other. If they are released at the same instant from rest, how fast will they be moving when they are very far away from each other? Assume they have iden­tical masses of 1.0 mg.

31. (II) An electron starts from rest 42.5 cm from a fixed point charge with Q = -0.125 nC. How fast will the electron be moving when it is very far away?

32. (II) Two equal but opposite charges are separated by a distance d, as shown in Fig. 23-28. Determine a formula for Vba = Vb _ Va for points B and A on the line between the charges situated as shown.

h HFIGURE 23-28Problem 32. +q B

23-4 Potential Due to Charge Distribution33. (II) A thin circular ring of radius R (as in Fig. 23-14) has

charge +Q/2 uniformly distributed on the top half, and —Ql2 on the bottom half, (a) What is the value of the electric potential at a point a distance x along the axis through the center of the circle? (b) What can you say about the electric field F, at a distance r alnna the axis? Let V = 0

34. (II) Three point charges are arranged at the comers of a square of side £ as shown in Fig. 23-29.What is the potential at the fourth comer (point A), taking V = 0 at a great distance?

FIGURE 23-29Problem 34.

-2Q

+3Q

j/?2

35. (II) A flat ring of inner radius Ri and outer radius R2, Fig. 23-30, carries a uniform surface charge density a. Determine the elec­tric potential at points along the axis *(the x axis). [Hint: Try substitutingvariables.] [' ̂ ;_____ v

I 1 1

FIGURE 23-30Problem 35.

36. (II) A total charge Q is uniformly distributed on a thread of length £. The thread forms a semicircle. What is the potential at the center? (Assume V = 0 at large distances.)

37. (II) A 12.0-cm-radius thin ring carries a uniformly distributed15.0 /jlC charge. A small 7.5-g sphere with a charge of 3.0 /jlC is placed exactly at the center of the ring and given a very small push so it moves along the ring axis (+x axis). How fast will the sphere be moving when it is 2.0 m from the center of the ring (ignore gravity)?

38. (II) A thin rod of length 2£ is centered on the x axis as shown in Fig. 23-31. The rod carries a uniformly distributed charge Q. Determine the potential V asa function of y for points along the y axis. Let V = 0 at infinity.

FIGURE 23-31Problems 38,39,40, and 53.

f21

39. (II) Determine the potential V(x) for points along the x axis outside the rod of Fig. 23-31 (Problem 38).

40. (Ill) The charge on the rod of Fig. 23-31 has a nonuni­form linear charge distribution, A = ax. Determine the potential V for (a) points along the y axis and (b) points along the x axis outside the rod.

41. (Ill) Suppose the flat circular disk of Fig. 23-15 (Example 23-9) has a nonuniform surface charge density a = ar2, where r is measured from the center of the disk. Find the potential V(x) at points along the x axis, relative to V = 0 at x = oo.

23-5 Equipotentials42. (I) Draw a conductor in the shape of a football. This

conductor carries a net negative charge, —Q. Draw in a dozen or so electric field lines and equipotential lines.

43. (II) Equipotential surfaces are to be drawn 100 V apart near a very large uniformly charged metal plate carrying a surface charge density cr = 0.75 /jlC/ui2. H o w far apart (in space) are the equipotential surfaces?

44. (II) A metal sphere of radius r0 = 0.44 m carries a charge Q = 0.50 fxC. Equipotential surfaces are to be drawn for 100-V intervals outside the sphere. Determine the radius r of(n\ the first (hi\ the tenth and (r \ the 100th enninntential

45. (II) Calculate the electric potential due to a tiny dipole whose dipole moment is 4.8 X IO-30 C • m at a point 4.1 X IO-9 m away if this point is (a) along the axis of the dipole nearer the positive charge; (b) 45° above the axis but nearer the positive charge; (c) 45° above the axis but nearer the negative charge. Let V = 0 at r = oo.

46. (Ill) The dipole moment, considered as a vector, points from the negative to the positive charge. The water molecule, Fig. 23-32, has a dipole moment p which can be considered as the vector sum of the two dipole moments p! and P2 as shown. The distance between each H and the O is about 0.96 X 10-10m; the lines joining the center of the O atom with each H atom make an angle of 104° as shown, and the net dipole moment has been measured to be p = 6.1 X 10_30C-m. (a) Determine the effective charge q on each H atom, (b) Determine the electric poten­tial, far from the molecule, due to each dipole, pi and p2, and show that

1 p cos 0V = ------ ---------,

47760 r2where p is the magnitude of the net dipole moment, p = pi + p2 , and V is the total potential due to both P) and p2. Take V = 0 at r = oo.

FIGURE 23-32Problem 46.

23-7 E Determined from V47. (I) Show that the electric field of a single point charge

(Eq. 21-4) follows from Eq. 23-5, V = (l/4i7e0)(g /r) .48. (I) What is the potential gradient just outside the surface of

a uranium nucleus (Q = +92e) whose diameter is about 15 X 10-15 m?

49. (II) The electric potential between two parallel plates is given by V (x ) = (8.0 V /m )* + 5.0 V, with x = 0 taken at one of the plates and x positive in the direction toward the other plate. What is the charge density on the plates?

50. (II) The electric potential in a region of space varies asV = by /(a2 + y2). Determine E.

51. (II) In a certain region of space, the electric potential is given by V = y2 + 2.5xy - 3.5jcyz. Determine the electric field vector, E, in this region.

52. (II) A dust particle with mass of 0.050 g and a charge of2.0 X 10-6 C is in a region of space where the potential is given by V(x) = (2.0 V /m 2)*2 - (3.0 V /m 3)*3. If the particle starts at x = 2.0 m, what is the initial acceleration of the charge?

53. (Ill) Use the results of Problems 38 and 39 to determine the electric field due to the uniformly charged rod of Fig. 23-31 for points (a) along the y axis and (b) along the x axis.

23-8 Electrostatic Potential Energy; Electron Volt54. (I) How much work must be done to bring three electrons

from a great distance apart to within 1.0 X 10-10m from one another (at the corners of an equilateral triangle)?

5 5 . W hat notential difference is needed to aive a helium

23-6 Dipoles 56. (I) What is the speed of (a) a 1.5-keV (kinetic energy) electron and (b) a 1.5-keV proton?

57. (II) Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.110 nm, and their final separation is 0.100 nm. How much electric potential energy was lost in this reaction (in units of eV)?

58. (II) An alpha particle (which is a helium nucleus, Q = +2e, m = 6.64 X 10-27 kg) is emitted in a radioactive decay with kinetic energy 5.53 MeV. What is its speed?

59. (II) Write the total electrostatic potential energy, U, for(a) four point charges and (b) five point charges. Draw a diagram defining all quantities.

60. (II) Four equal point charges, Q, are fixed at the corners of a square of side b. (a) What is their total electrostatic poten­tial energy? (b) How much potential energy will a fifth charge, Q, have at the center of the square (relative to V = 0 at r = oo)? (c) If constrained to remain in that plane, is the fifth charge in stable or unstable equilibrium? If unstable, what maximum kinetic energy could it acquire? (d) If a nega­tive ( - Q ) charge is at the center, is it in stable equilibrium?

61. (II) An electron starting from rest acquires 1.33 keV of kinetic energy in moving from point A to point B. (a) How much kinetic energy would a proton acquire, starting from rest at B and moving to point A? (b) Determine the ratio of their speeds at the end of their respective trajectories.

62. (II) Determine the total electrostatic potential energy of a conducting sphere of radius r0 that carries a total charge Q distributed uniformly on its surface.

63. (II) The liquid-drop model of the nucleus suggests that high- energy oscillations of certain nuclei can split (“fission”) a large nucleus into two unequal fragments plus a few neutrons. Using this model, consider the case of a uranium nucleus fissioning into two spherical fragments, one with a charge qi = +38e and radius r\ = 5.5 X 10-15m, the other with q2 = +54e and r2 = 6.2 X 10-15m. Calculate the electric potential energy (MeV) of these fragments, assuming that the charge is uniformly distributed throughout the volume of each spherical nucleus and that their surfaces are initially in contact at rest. The electrons surrounding the nuclei can be neglected. This electric potential energy will then be entirely converted to kinetic energy as the fragments repel each other. How does your predicted kinetic energy of the fragments agree with the observed value associated with uranium fission (approxi­mately 200 MeV total)? [l MeV = 106eV.]

64. (Ill) Determine the total electrostatic potential energy of a nonconducting sphere of radius r0 carrying a total charge Q distributed uniformly throughout its volume.

*23-9 CRT*65. (I) Use the ideal gas as a model to estimate the rms speed

of a free electron in a metal at 273 K, and at 2700 K (a typical temperature of the cathode in a CRT).

* 66. (Ill) Electrons are accelerated by 6.0 kV in a CRT. The screenis 28 cm wide and is 34 cm from the 2.6-cm-long deflection plates. Over what range must the horizontally deflecting elec­tric field vary to sweep the beam fully across the screen?

* 67. (Ill) In a given CRT, electrons are accelerated horizontallyby 7.2 kV. They then pass through a uniform electric field E for a distance of 2.8 cm which deflects them upward so they reach the tnn n f the screen cm awav. 11 cm abnve the

| General Problems68. If the electrons in a single raindrop, 3.5 mm in diameter, could

be removed from the Earth (without removing the atomic nuclei), by how much would the potential of the Earth increase?

69. By rubbing a nonconducting material, a charge of 10-8 C can readily be produced. If this is done to a sphere of radius 15 cm, estimate the potential produced at the surface. Let V = 0 at r = oo.

70. Sketch the electric field and equipotential lines for two charges of the same sign and magnitude separated by a distance d.A +33 fxC point charge is placed 36 cm from an identical +33 fxC charge. A —1.5 /xC charge is moved from point a to point b, Fig. 23-33. ^What is the change •

71

in potential energy?

FIGURE 23-33Problem 71.

33 iiC( 12 cm14 cm

24 cma

+33 [xC

72. At each corner of a cube of side £ there is a point charge Q, Fig. 23-34. (a) What is the potential at the center of the cube (V = 0 at r = oo)? (b) What is the potential at each corner due to the other seven charges? (c) What is the total potential energy of this system?

FIGURE 23-34Problem 72.

73. In a television picture tube (CRT), electrons are accelerated by thousands of volts through a vacuum. If a television set is laid on its back, would electrons be able to move upward against the force of gravity? What potential difference, acting over a distance of 3.5 cm, would be needed to balance the downward force of gravity so that an electron would remain stationary? Assume that the electric field is uniform.

74. Four point charges are located at the corners of a square that is 8.0 cm on a side. The charges, going in rotation around the square, are Q, 2Q, —3Q, and 2Q, where Q = 3.1 ixC (Fig. 23-35). What is the total electric potential energy stored in the system, relative to U = 0 at infinite separation?

FIGURE 23-35Problem 74. 2Q

75. In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. See Fig. 23-36. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. If the plate voltage is -3.02 V (compared to the barium) when the fastest elec­trons are stopped, what was the speed of these electrons when they were emitted?

Plate

UV

FIGURE 2 3 -3 6

■y =-3.02 V

76. An electron is accelerated horizontally from rest in a televi­sion picture tube by a potential difference of 5500 V. It then passes between two horizontal plates 6.5 cm long and 1.3 cm apart that have a potential difference of 250 V (Fig. 23-37). At what angle 0 will the electron be traveling after it passes between the plates?

FIGURE 23-37Problem 76.

] 0

77. Three charges are at the corners of an equilateral triangle (side £) as shown in Fig. 23-38.Determine the potential at the ymidpoint of each of the sides. Let V = 0 at r = oo.

FIGURE 23-38Problem 77.

78. Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass m = 0.340 kg are dropped from a height of 2.00 m, but one of the balls is positively charged with qi = 450 fxC, and the second is negatively charged with q2 = —450/xC . Use conservation of energy to determine the difference in the speeds of the two balls when they hit the ground. (Neglect air resistance.)

79. A lightning flash transfers 4.0 C of charge and 4.8 MJ of energy to the Earth, (a) Between what potential difference did it travel? (b) How much water could this energy boil, starting from room temperature? [Hint: See Chapter 19.]

80. Determine the components of the electric field, Ex and Ey , as a function of x and y inthe xy plane due to a dipole, Fig. 23-39, starting with Eq. 23-7. Assume r = (x2 + y2)l » I

FIGURE 23-39Problem 80. Q h^£-H +Q

V=0

81. A nonconducting sphere of radius r2 contains a concentric spherical cavity of radius r\ . The material between r\ and r2 carries a uniform charge density pE (C/m3). Determine the electric potential V, relative to V = 0 at r = oo, as a function of the distance r from the center for (a) r > r2, (P) r\ < r < r2> and (c) 0 < r < ri. Is V continuous at rx and r2l

82. A thin flat nonconducting disk, with radius R0 and charge Q, has a hole with a radius R0/2 in its center. Find the elec­tric potential V (*) at points along the symmetry (*) axis of the disk fa line nernendicular to the disk, nassinp through its

83. A Geiger counter is used to detect charged particles emitted by radioactive nuclei. It consists of a thin, positively charged central wire of radius Ra surrounded by a concentric conducting cylinder of radius Rb with an equal negative charge (Fig. 23-40). The charge per unit length on the inner wire is A (units C/m). The interior space between wire and cylinder is filled with low-pressure inert gas. Charged particles ionize some of these gas atoms; the resulting free electrons are attracted toward the positive central wire. If the radial electric field is strong enough, the freed electrons gain enough energy to ionize other atoms, causing an “avalanche” of electrons to strike the central wire, generating an electric “signal.” Find the expression for the electric field between the wire and the cylinder, and show that the potential difference between Ra and Rb is

Va ~ Vh = ( AV 2'77€<

1 I Rb ln ' ^

-"h

FIGURE 23-40Problem 83.

Central wire, radius !<.

84. A Van de Graaff generator (Fig. 23-41) can develop a very large potential difference, even millions of volts. Electrons are pulled off the belt by the high voltage pointed electrode at A, leaving the belt positively charged. (Recall Example 23-5 where we saw that near sharp points the electric field is high and ionization can occur.) The belt carries the positive charge up inside the spherical shell where electrons from the large conducting sphere are attracted over to the pointed conductor at B, leaving the outer surface of the conducting sphere positively charged. As more charge is brought up, the sphere reaches extremely high voltage. Consider a Van de Graaff generator with asphere of radius 0.20 m. (a) What is the electric potential on the surface of the sphere when electrical breakdown occurs? (Assume V = 0 at r = oo.) (b) What is the charge on the sphere for the poten­tial found in part (a)?

GENERATOR+ + + Conductor

V+(___ tut __+ (^PuUev

I+ P

+ 1

FIGURE 23-41Problem 84.

50 kV JT| aW I * —

Insulator

► Motor*driven 1 pulley

85. The potential in a region of space is given by V = B/(x2 + R2)2 where b = 150 V • m4 and R = 0.20 m. (a) Find V at x = 0.20 m. (b) Find E as a function of x. (c) Find E at x = 0.20 m.

86. A charge — q\ of mass m rests on the y axis at a distance b above the x axis. Two positive charges of magnitude +q2 are fixed on the x axis at x = +a and x = — a, respectively (Fig. 23-42). If the —qi charge is given an initial velocity vQ in the positive y direction, what is the minimum value of v0 such that the chargeescapes to a point infi­nitely far away from the two positive charges?

+b

FIGURE 23-42Problem 86.

+#2

-q i , m

+<l2-a 0 +a

* Numerical/Computer*87. (II) A dipole is composed of a —1.0 nC charge at

x = -1.0 cm and a +1.0 nC charge at x = +1.0 cm. (a) Make a plot of V along the x axis from x = 2.0 cm to x = 15 cm. (b) On the same graph, plot the approximate V using Eq. 23-7 from x = 2.0 cm to x = 15 cm. Let V = 0 at x = oo.

* 88. (II) A thin flat disk of radius R0 carries a total charge Q thatis distributed uniformly over its surface. The electric poten­tial at a distance x on the x axis is given by

QV(x) = [(*2 + R i f - 42lT€0i?0

(See Example 23-9.) Show that the electric field at a distance x on the x axis is given by

QE(x) =2ire0Rt

1 -

(x2 + R H

Make graphs of V(x) and E(x) as a function of x /R 0 for x /R 0 = 0 to 4. (Do the calculations in steps of 0.1.) Use Q = 5.0 fxC and R0 = 10 cm for the calculation and graphs.

*89. (Ill) You are trying to determine an unknown amount of charge using only a voltmeter and a ruler, knowing that it is either a single sheet of charge or a point charge that is creating it. You determine the direction of greatest change of potential, and then measure potentials along a line in that direction. The potential versus position (note that the zero of position is arbitrary, and the potential is measured rela­tive to ground) is measured as follows:

x (cm) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 V (volts) 3.9 3.0 2.5 2.0 1.7 1.5 1.4 1.4 1.2 1.1

(a) Graph V versus position. Do you think the field is caused by a sheet or a point charge? (b) Graph the data in such a way that you can determine the magnitude of the charge and determine that value, (c) Is it possible to determine where the charge is from this data? If so, give the position of the charge.

Answers to Exercises

A: (b). D: (a) iii, (b) i, (c) i.B: (d). E: (c).