Q e ncl

€0

Gauss's LawCHAPTER-OPENING QUESTION—Guess now!A nonconducting sphere has a uniform charge density throughout. How does the magnitude of the electric field vary inside with distance from the center?

(a) The electric field is zero throughout.(b) The electric field is constant but nonzero throughout.(c) The electric field is linearly increasing from the center to the outer edge.(d) The electric field is exponentially increasing from the center to the outer edge.(e) The electric field increases quadratically from the center to the outer edge.

The great mathematician Karl Friedrich Gauss (1777-1855) developed an important relation, now known as Gauss’s law, which we develop and discuss in this Chapter. It is a statement of the relation between electric charge and electric field and is a more general and elegant form of Coulomb’s law.

We can, in principle, determine the electric field due to any given distribution of electric charge using Coulomb’s law. The total electric field at any point will be the vector sum (or integral) of contributions from all charges present (see Eq. 21-6). Except for some simple cases, the sum or integral can be quite complicated to evaluate. For situations in which an analytic solution (such as we carried out in the Examples of Sections 21-6 and 21-7) is not possible, a computer can be used.

In some cases, however, the electric field due to a given charge distribution can be calculated more easily or more elegantly using Gauss’s law, as we shall see in this Chapter. But the maior importance of Gauss’s law is that it gives us additional

Gauss’s law is an elegant relation between electric charge and electric field. It is more general than Coulomb’s law. Gauss’s law involves an integral of the electric field E at each point on a closed surface. The surface is only imaginary, and we choose the shape and placement of the surface so that we can evaluate the integral. In this drawing, two different 3-D surfaces are shown (one green, one blue), both enclosing a point charge Q. Gauss’s law states that the product E • dA , where d A is an infinitesimal area of the surface, integrated over the entire surface, equals the charge enclosed by the surface <2enci divided by e0. Both surfaces here enclose the same charge Q. Hence <f>E • d A will give the same result for both surfaces.

T * * 22

CONTENTS22-1 Electric Flux2 2-2 Gauss’s Law22-3 Applications of Gauss’s

Law*22-4 Experimental Basis

of Gauss’s and Coulomb’s Laws

FIGURE 22-1 (a) A uniform electric field E passing through a flat area A. (b) E± = E cos 0 is the component of E perpendicular to the plane of area A. (c) A ± = A cos 6 is the projection (dashed) of the area A perpendicular to the field E.

FIGURE 22-2 Electric flux through a curved surface. One small area of the surface, A A /, is indicated.

FIGURE 22-3 Electric flux through a closed surface.

2 2 —1 Electric FluxGauss’s law involves the concept of electric flux, which refers to the electric field passing through a given area. For a uniform electric field E passing through an area A, as shown in Fig. 22-la, the electric flux <&E is defined as

<PE = E A cos 6,where 0 is the angle between the electric field direction and a line drawn perpendicular to the area. The flux can be written equivalently as

<E>£ = El A = E A ± = EA cos 6, [E uniform] (22-la)

where E± = E cos 0 is the component of E along the perpendicular to the area (Fig. 22-lb) and, similarly, A ± = A cos 0 is the projection of the area A perpen­dicular to the field E (Fig. 22-lc).

The area A of a surface can be represented by a vector A whose magnitude is A and whose direction is perpendicular to the surface, as shown in Fig. 22-lc. The angle 0 is the angle between E and A, so the electric flux can also be written

<I>£ = E • A. [E uniform] (22-lb)Electric flux has a simple intuitive interpretation in terms of field lines. We

mentioned in Section 21-8 that field lines can always be drawn so that the number (N) passing through unit area perpendicular to the field (A±) is proportional to the magnitude of the field (E): that is, E oc N /A ±. Hence,

N oc E A ± =so the flux through an area is proportional to the number of lines passing through that area.

Electric flux. Calculate the electric flux through the rectangle shown in Fig. 22-la. The rectangle is 10 cm by 20 cm, the electric field is uniform at 200 N/C, and the angle 0 is 30°.APPROACH We use the definition of flux, = E • A = E A cos 0.SOLUTION The electric flux is

<$>E = (200 N/C) (0.10 m X 0.20 m) cos 30° = 3.5N-m2/C.

EXERCISE A Which of the following would cause a change in the electric flux through a circle lying in the xz plane where the electric field is (10 N) j ? (a) Changing the magnitude of the electric field, (b) Changing the size of the circle, (c) Tipping the circle so that it is lying in the xy plane. (d) All of the above, (e) None of the above.

In the more general case, when the electric field E is not uniform and the surface is not flat, Fig. 22-2, we divide up the chosen surface into n small elements of surface whose areas are AA 1, AA2, • • •, AA n. We choose the division so that each AA t is small enough that (1) it can be considered flat, and (2) the electric field varies so little over this small area that it can be considered uniform. Then the electric flux through the entire surface is approximately

<&£ « S E r A i , . ,i=1

where Ef is the field passing through AAf. In the limit as we let AAj —» 0, the sum becomes an integral over the entire surface and the relation becomes mathematically exact:

<DE = | E • dA. (22-2)

Gauss’s law involves the total flux through a closed surface—a surface of any shape that completely encloses a volume of space, such as that shown in Fig. 22-3. In this case, the net flux through the enclosing surface is given by

EXAMPLE 22-1

Up to now we have not been concerned with an ambiguity in the direction of the vector A or dA that represents a surface. For example, in Fig. 22-lc, the vector A could point upward and to the right (as shown) or downward to the left and still be perpendicular to the surface. For a closed surface, we define (arbitrarily) the direction of A, or of dA, to point outward from the enclosed volume, Fig. 22-4. For an electric field line leaving the enclosed volume (on the right in Fig. 22-4), the angle 6 between E and dA must be less than tt /2(= 90°); hence cos0 > 0. For a line entering the volume (on the left in Fig. 22-4) 6 > it/2; hence cos 6 < 0. Hence, flux entering the enclosed volume is negative ( fE cos 6 dA < 0), whereas flux leaving the volume is positive. Consequently, Eq. 22-3 gives the net flux out of the volume. If 4>£ is negative, there is a net flux into the volume.

In Figs. 22-3 and 22-4, each field line that enters the volume also leaves the volume. Hence <E>£ = $E • dA = 0. There is no net flux into or out of this enclosed surface. The flux, <J>E • dA, will be nonzero only if one or more lines start or end within the surface. Since electric field lines start and stop only on electric charges, the flux will be nonzero only if the surface encloses a net charge. For example, the surface labeled A 1 in Fig. 22-5 encloses a positive charge and there is a net outward flux through this surface (<&£ > 0). The surface A 2 encloses an equal magnitude negative charge and there is a net inward flux (<&£ < 0). For the configuration shown in Fig. 22-6, the flux through the surface shown is negative (count the lines). The value of 4>£ depends on the charge enclosed by the surface, and this is what Gauss’s law is all about.

2 2 -2 Gauss's LawThe precise relation between the electric flux through a closed surface and the net charge <2encl enclosed within that surface is given by Gauss’s law:

Qe nclE - d A = (22-4)

where e0 is the same constant (permittivity of free space) that appears in Coulomb’s law. The integral on the left is over the value of E on any closed surface, and we choose that surface for our convenience in any given situation. The charge Qenci is the net charge enclosed by that surface. It doesn’t matter where or how the charge is distributed within the surface. Any charge outside this surface must not be included. A charge outside the chosen surface may affect the position of the electric field lines, but will not affect the net number of lines entering or leaving the surface. For example, Qend for the gaussian surface A 1 in Fig. 22-5 would be the positive charge enclosed by A x; the negative charge does contribute to the electric field at A 1 but it is not enclosed by surface A 1 and so is not included in Qenc\ .

Now let us see how Gauss’s law is related to Coulomb’s law. First, we show that Coulomb’s law follows from Gauss’s law. In Fig. 22-7 we have a single isolated charge Q. For our “gaussian surface,” we choose an imaginary sphere of radius r centered on the charge. Because Gauss’s law is supposed to be valid for any surface, we have chosen one that will make our calculation easy. Because of the symmetry of this (imaginary) sphere about the charge at its center, we know that E must have the same magnitude at any point on the surface, and that E points radially outward (inward for a negative charge) parallel to dA, an element of the surface area. Hence, we write the integral in Gauss’s law as

i E d A = E ^d A = E(4nr2)

since the surface area of a sphere of radius r is 4irr2, and the magnitude of E is the same at all points on this spherical surface. Then Gauss’s law becomes, with Qencl = Q,

Qe0

because E and dA are both perpendicular to the surface at each point, and cos 6 = 1. Solving for E we obtain

E - d A =

— = cbE-dA = E{A7rr7

FIGURE 22-4 The direction of an element of area dA is taken to point outward from an enclosed surface.

FIGURE 22-5 An electric dipole. Flux through surface A \ is positive. Flux through A 2 is negative.

FIGURE 22-6 Net flux through surface A is negative.

FIGURE 22-7 A single point charge Q at the center of an imaginary sphere of radius r (our “gaussian surface”—that is, the closed surface we choose to use for applying Gauss’s law in this case).

FIGURE 22-8 A single point charge surrounded by a spherical surface, A 1, and an irregular surface, A 2.

FIGURE 22-9 Electric flux through a closed surface. (Same as Fig. 22-3.) No electric charge is enclosed by this surface (Qencl = 0 ).

Now let us do the reverse, and derive Gauss’s law from Coulomb’s law for static electric charges1. First we consider a single point charge Q surrounded by an imaginary spherical surface as in Fig. 22-7(and shown again, green, in Fig. 22-8). Coulomb’s law tells us that the electric field at the spherical surface is E = (l/477e0)(<2A2)- Reversing the argument we just used, we have

E • dA = 1 QdA Q(47rr) = — •

4-7760 r2 477€0r2This is Gauss’s law, with Qend = Q, and we derived it for the special case of a spherical surface enclosing a point charge at its center. But what about some other surface, such as the irregular surface labeled A 2 in Fig. 22-8? The same number of field lines (due to our charge Q) pass through surface A 2, as pass through the spherical surface, A x. Therefore, because the flux through a surface is proportional to the number of lines through it as we saw in Section 22-1, the flux through A 2 is the same as through A x:

E • dA =

Hence, we can expect that

E d A = —«o

E dA = — e0

E; • dA = — >

would be valid for any surface surrounding a single point charge Q.Finally, let us look at the case of more than one charge. For each charge, Qt,

enclosed by the chosen surface,ae o

where Ef refers to the electric field produced by Qt alone. By the superposition principle for electric fields (Section 21-6), the total field E is equal to the sum of the fields due to each separate charge, E = DEf. Hence

Q enclE • dA = dA =

where Qenc\ = 2 Qt is the total net charge enclosed within the surface. Thus we see, based on this simple argument, that Gauss’s law follows from Coulomb’s law for any distribution of static electric charge enclosed within a closed surface of any shape.

The derivation of Gauss’s law from Coulomb’s law is valid for electric fields produced by static electric charges. We will see later that electric fields can also be produced by changing magnetic fields. Coulomb’s law cannot be used to describe such electric fields. But Gauss’s law is found to hold also for electric fields produced in any of these ways. Hence Gauss’s law is a more general law than Coulomb’s law. It holds for any electric field whatsoever.

Even for the case of static electric fields that we are considering in this Chapter, it is important to recognize that E on the left side of Gauss’s law is not necessarily due only to the charge Qend that appears on the right. For example, in Fig. 22-9 there is an electric field E at all points on the imaginary gaussian surface, but it is not due to the charge enclosed by the surface (which is Qenc\ = 0 in this case). The electric field E which appears on the left side of Gauss’s law is the total electric field at each point, on the gaussian surface chosen, not just that due to the charge 0 encl, which appears on the right side. Gauss’s law has been found to be valid for the total field at any surface. It tells us that any difference between the input and output flux of the electric field over any surface is due to charge within that surface.'''Note that Gauss’s law would look more complicated in terms of the constant k = l/47re0 that we originally used in Coulomb’s law (Eqs. 21-1 or 21-4a):

Coulomb’s law Gauss’s law

E • dA = 4irkQ

CONCEPTUAL EXAMPLE 22-21 Flux from Gauss's law. Consider the twogaussian surfaces, A 1 and A 2, shown in Fig. 22-10. The only charge present is the charge Q at the center of surface A x. What is the net flux through each surface, A 1 and A 2 ? RESPONSE The surface A 1 encloses the charge +Q. By Gauss’s law, the net flux through A 1 is then Q /e0. For surface A 2, the charge +Q is outside the surface. Surface A 2 encloses zero net charge, so the net electric flux through A 2 is zero, by Gauss’s law. Note that all field lines that enter the volume enclosed by surface A 2 also leave it.

EXERCISE B A point charge Q is at the center of a spherical gaussian surface A . When a second charge Q is placed just outside A , the total flux through this spherical surface A is(a) unchanged, (b) doubled, (c) halved, (d) none of these.

EXERCISE C Three 2.95 /xC charges are in a small box. What is the net flux leaving the box? (a) 3.3 X 1012N-m 2/C, (b) 3.3 X 105N-m 2/C, (c) 1.0 X 1012N-m 2/C,(d) 1.0 X 106 N • m2/C, (e) 6.7 X 106N-m 2/C.

We note that the integral in Gauss’s law is often rather difficult to carry out in practice. We rarely need to do it except for some fairly simple situations that we now discuss.

2 2 -3 Applications o f Gauss's LawGauss’s law is a very compact and elegant way to write the relation between electric charge and electric field. It also offers a simple way to determine the electric field when the charge distribution is simple and/or possesses a high degree of symmetry. In order to apply Gauss’s law, however, we must choose the “gaussian” surface very carefully (for the integral on the left side of Gauss’s law) so we can determine E. We normally try to think of a surface that has just the symmetry needed so that E will be constant on all or on parts of its surface. Sometimes we choose a surface so the flux through part of the surface is zero.( 2 5 E E H S H B Spherical conductor. A thin spherical shell of radius r0

possesses a total net charge Q that is uniformly distributed on it (Fig. 22-11). Determine the electric field at points (a) outside the shell, and (b) inside the shell, (c) What if the conductor were a solid sphere?APPROACH Because the charge is distributed symmetrically, the electric field must also be symmetric. Thus the field outside the sphere must be directed radially outward (inward if Q < 0) and must depend only on r, not on angle (spherical coordinates). SOLUTION (a) The electric field will have the same magnitude at all points on an imaginary gaussian surface, if we choose that surface as a sphere of radius r (r > r0) concentric with the shell, and shown in Fig. 22-11 as the dashed circle A 1. Because E is perpendicular to this surface, the cosine of the angle between E and dA is always 1. Gauss’s law then gives (with Send = Q in Eq. 22-4)

E • dA = £(4irr2) = —, e0

where 4irr2 is the surface area of our sphere (gaussian surface) of radius r. Thus

E = 1 Q> r0]4t7€0 r2

Thus the field outside a uniformly charged spherical shell is the same as if all the charge were concentrated at the center as a point charge.(b) Inside the shell, the electric field must also be symmetric. So E must again have the same value at all points on a spherical gaussian surface (A2 in Fig. 22-11) concentric with the shell. Thus E can be factored out of the integral and, with Genci = 0 because the charge enclosed within the sphere A 2 is zero, we have

Hence

E - d A = £(4tjt2) = 0.

E = 0

FIGURE 22-10 Example 22-2. Two gaussian surfaces.

FIGURE 22-11 Cross-sectional drawing of a thin spherical shell of radius r0, carrying a net charge Q uniformly distributed. A 1 and A 2 represent two gaussian surfaces we use to determine E. Example 22-3.

\ r < r0l

FIGURE 22-12 A solid sphere of uniform charge density. Example 22-4.

FIGURE 22-13 Magnitude of the electric field as a function of the distance r from the center of a uniformly charged solid sphere.

EXERCISE D A charge Q is placed on a hollow metal ball. We saw in Chapter 21 that the charge is all on the surface of the ball because metal is a conductor. How does the charge distribute itself on the ball? (a) Half on the inside surface and half on the outside surface, (ib) Part on each surface in inverse proportion to the two radii, (c) Part on each surface but with a more complicated dependence on the radii than in answer (b). (d) All on the inside surface, (e) All on the outside surface.

EXAMPLE 22-4 Solid sphere of charge. An electric charge Q is distributed uniformly throughout a nonconducting sphere of radius r0, Fig. 22-12. Determine the electric field (a) outside the sphere (r > rQ) and (b) inside the sphere (r < r0).APPROACH Since the charge is distributed symmetrically in the sphere, the electric field at all points must again be symmetric. E depends only on r and is directed radially outward (or inward if Q < 0).SOLUTION (a) For our gaussian surface we choose a sphere of radius r(r > r0), labeled A x in Fig. 22-12. Since E depends only on r, Gauss’s law gives, withGencl — Q ’

>E • d\ = £(4-7tt?' Qor

1 Q4/7T€0 r2

Again, the field outside a spherically symmetric distribution of charge is the same as that for a point charge of the same magnitude located at the center of the sphere.(b) Inside the sphere, we choose for our gaussian surface a concentric sphere of radius r { r < r0], labeled A 2 in Fig. 22-12. From symmetry, the magnitude of E is the same at all points on A 2 , and E is perpendicular to the surface, so

E • dA = E{Airr2).

We must equate this to <2enci/€o where Qencl is the charge enclosed by A 2 . <2encl is not the total charge Q but only a portion of it. We define the charge density, pE, as the charge per unit volume (pE = dQ/dV), and here we are given that pE = constant. So the charge enclosed by the gaussian surface A 2 , a sphere of radius r, is

( ̂ Trr3pE \ r3Gene, = f “ FT G = 3 f t

\3 irr0PE/ r0Hence, from Gauss’s law,

E(A7rr1) = = rlJ L*o rl €0

or

E = T ~ ^3 r - V < r o]4 ir e 0 r i

Thus the field increases linearly with r, until r = r0. It then decreases as 1/r2, as plotted in Fig. 22-13.

I EXERCISE E Return to the Chapter-Opening Question, page 591, and answer it again now. Try to explain why you may have answered differently the first time.

The results in Example 22-4 would have been difficult to obtain from Coulomb’s law by integrating over the sphere. Using Gauss’s law and the symmetry of the situation, this result is obtained rather easily, and shows the great power of Gauss’s law. However, its use in this way is limited mainly to cases where the charge distribution has a high degree of symmetry. In such cases, we choose a simple surface on which E = constant, so the integration is simple. Gauss’s law holds, of

EXAMPLE 22-5 Nonuniformly charged solid sphere. Suppose the charge density of the solid sphere in Fig. 22-12, Example 22-4, is given by pE = ar2, where a is a constant, (a) Find a in terms of the total charge Q on the sphere and its radius r0. (b) Find the electric field as a function of r inside the sphere.APPROACH We divide the sphere up into concentric thin shells of thickness dr as shown in Fig. 22-14, and integrate (a) setting Q = JpE dV and (b) using Gauss’s law. SOLUTION (a) A thin shell of radius r and thickness dr (Fig. 22-14) has volume dV = 4irr2 dr. The total charge is given by

Attol cQ = f pE dV = f (ar2)(A7rr2 dr) = A tto l fJ Jo Jo

r*dr =i o 5

Thus a = 5Q/A7rrQ.(b) To find E inside the sphere at distance r from its center, we apply Gauss’s law to an imaginary sphere of radius r which will enclose a charge

5< 2Q encl [ pE dV = [ (ar2) Airr2 dr = [

Jo Jo Jo 4777*0r Airr dr „ r 5

Q - 5 -roBy symmetry, E will be the same at all points on the surface of a sphere of radius r, so Gauss’s law gives

GenclE • dA =

( £ ) ( 4 i7 T 2) = Q ~ r~ 3 e0r0

SO

E =Q r

4ire0rg

FIGURE 22-14 Example 22-5.

EXAMPLE 22-6 Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length, A. Calculate the electric field at points near (but outside) the wire, far from the ends.APPROACH Because of the symmetry, we expect the field to be directed radially outward and to depend only on the perpendicular distance, R, from the wire. Because of the cylindrical symmetry, the field will be the same at all points on a gaussian surface that is a cylinder with the wire along its axis, Fig. 22-15. E is perpendicular to this surface at all points. For Gauss’s law, we need a closed surface, so we include the flat ends of the cylinder. Since E is parallel to the ends, there is no flux through the ends (the cosine of the angle between E and dA on the ends is cos 90° = 0).SOLUTION For our chosen gaussian surface Gauss’s law gives

Gencl =to ’

E • dA = E(2ttR£)*o

where I is the length of our chosen gaussian surface (£ « length of wire), and 2ttR is its circumference. Hence

1 AE =2776 n R

NOTE This is the same result we found in Example 21-11 using Coulomb’s law (we used x there instead of R), but here it took much less effort. Again we see the great power of Gauss’s law.fNOTE Recall from Chapter 10, Fig. 10-2, that we use R for the distance of a particle from an axis (cylindrical symmetry), but lower case r for the distance from a point (usually the origin 0).

FIGURE 22-15 Calculation of E due to a very long line of charge. Example 22-6.

1

'V sit

%i

i -i i - ■ » * * f +,i *% t

it/

J? ole,-, f™- o

FIGURE 22-16 Calculation of the electric field outside a large uniformly charged nonconducting plane surface. Example 22-7.

FIGURE 22-17 Electric field near surface of a conductor. Example 22-8.

/ t\ CAUTIONWhen is E = o"/e0 and

when is E = a / 2e0

EXAMPLE 22-7 Infinite plane of charge. Charge is distributed uniformly, with a surface charge density cr (cr = charge per unit area = dQ/dA), over a very large but very thin nonconducting flat plane surface. Determine the electric field at points near the plane.APPROACH We choose as our gaussian surface a small closed cylinder whose axis is perpendicular to the plane and which extends through the plane as shown in Fig. 22-16. Because of the symmetry, we expect E to be directed perpendicular to the plane on both sides as shown, and to be uniform over the end caps of the cylinder, each of whose area is A.SOLUTION Since no flux passes through the curved sides of our chosen cylindrical surface, all the flux is through the two end caps. So Gauss’s law gives

Qencl O -AE - d A = 2 EA =

e0 e0where Qencl = a A is the charge enclosed by our gaussian cylinder. The electric field is then

NOTE This is the same result we obtained much more laboriously in Chapter 21, Eq. 21-7. The field is uniform for points far from the ends of the plane, and close to its surface.

EXAMPLE 22-8 Electric field near any conducting surface. Show that the electric field just outside the surface of any good conductor of arbitrary shape is given by

E = — > e o

where cr is the surface charge density on the conductor’s surface at that point.APPROACH We choose as our gaussian surface a small cylindrical box, as we did in the previous Example. We choose the cylinder to be very small in height, so that one of its circular ends is just above the conductor (Fig. 22-17). The other end is just below the conductor’s surface, and the sides are perpen­dicular to it.SOLUTION The electric field is zero inside a conductor and is perpendicular to the surface just outside it (Section 21-9), so electric flux passes only through the outside end of our cylindrical box; no flux passes through the short sides or inside end. We choose the area A (of the flat cylinder end) small enough so that E is essentially uniform over it. Then Gauss’s law gives

Q encl AE - d A = EAe o

so that

E = — • [at surface of conductor] (22-5)€o

NOTE This useful result applies for a conductor of any shape.

Why is it that the field outside a large plane nonconductor is E = cr/2e0 (Example 22-7) whereas outside a conductor it is E = o-/e0 (Example 22-8)? The reason for the factor of 2 comes not from conductor verses nonconductor. It comes instead from how we define charge per unit area cr. For a thin flat noncon­ductor, Fig. 22-16, the charge may be distributed throughout the volume (not only on the surface, as for a conductor). The charge per unit area cr represents all the charge throughout the thickness of the thin nonconductor. Also our gaussian surface has its

For a conductor, on the other hand, the charge accumulates on the outer surfaces only. For a thin flat conductor, as shown in Fig. 22-18, the charge accumulates on both surfaces, and using the same small gaussian surface we did in Fig. 22-17, with one end inside and the other end outside the conductor, we came up with the result, E = cr/e0. If we defined cr for a conductor, as we did for a nonconductor, cr would represent the charge per area for the entire conductor. Then Fig. 22-18 would show 0-/2 as the surface charge on each surface, and Gauss’s law would give JE • dA = EA = (o-/2)A /e0 = crA/2e0 so E = just as for anonconductor. We need to be careful about how we define charge per unit area a.

We saw in Section 21-9 that in the static situation, the electric field inside any conductor must be zero even if it has a net charge. (Otherwise, the free charges in the conductor would move—until the net force on each, and hence E, were zero.) We also mentioned there that any net electric charge on a conductor must all reside on its outer surface. This is readily shown using Gauss’s law. Consider any charged conductor of any shape, such as that shown in Fig. 22-19, which carries a net charge Q. We choose the gaussian surface, shown dashed in the diagram, so that it all lies just below the surface of the conductor and encloses essentially the whole volume of the conductor. Our gaussian surface can be arbitrarily close to the surface, but still inside the conductor. The electric field is zero at all points on this gaussian surface since it is inside the conductor. Hence, from Gauss’s law, Eq. 22-4, the net charge within the surface must be zero. Thus, there can be no net charge within the conductor. Any net charge must lie on the surface of the conductor.

If there is an empty cavity inside a conductor, can charge accumulate on that (inner) surface too? As shown in Fig. 22-20, if we imagine a gaussian surface (shown dashed) just inside the conductor above the cavity, we know that E must be zero everywhere on this surface since it is inside the conductor. Hence, by Gauss’s law, there can be no net charge at the surface o f the cavity.

But what if the cavity is not empty and there is a charge inside it?

+ +

+ +cr cr

\ + + /

+ +

I" ■+ i +

ili■ w

i+ i +

11

E - E+ +

+ +

FIGURE 22-18 Thin flat charged conductor with surface charge density <r at each surface. For the conductor as a whole, the charge density is cr' = 2cr.

FIGURE 22-19 An insulated charged conductor of arbitrary shape, showing a gaussian surface (dashed) just below the surface of the conductor.

FIGURE 22-20 An empty cavity inside a charged conductor carries zero net charge.

FIGURE 22-21 Example 22-9.

CONCEPTUAL EXAMPLE 22-51 Conductor with charge inside a cavity.Suppose a conductor carries a net charge +Q and contains a cavity, inside of which resides a point charge +q. What can you say about the charges on the inner and outer surfaces of the conductor?

RESPONSE As shown in Fig. 22-21, a gaussian surface just inside the conductor surrounding the cavity must contain zero net charge (E = 0 in a conductor). Thus a net charge of —q must exist on the cavity surface. The conductor itself carries a net charge +Q, so its outer surface must carry a charge equal

O

V05

*S ° L V /

EXERCISE F Which of the following statements about Gauss’s law is correct? (a) If we know the charge enclosed by a surface, we always know the electric field everywhere at the surface.(b) When finding the electric field with Gauss’s law, we always use a sphere for the gaussian surface.(c) If we know the total flux through a surface, we also know the total charge inside the surface, (id) We can only use Gauss’s law if the electric field is constant in space.

Gauss's Law for Symmetric Charge Distributions1. First identify the symmetry of the charge distribu­

tion: spherical, cylindrical, planar. This identification should suggest a gaussian surface for which E will be constant and/or zero on all or on parts of the surface: a sphere for spherical symmetry, a cylinder for cylindrical symmetry and a small cylinder or “pillbox” for planar symmetry.

2. Draw the appropriate gaussian surface making sure it passes through the point where you want to know the electric field.

3. Use the symmetry of the charge distribution to determine the direction of E at points on the gaussian surface.

4. Evaluate the flux, <f)E • dA. With an appropriate gaussian surface, the dot product E • dA should be zero or equal to + E dA, with the magnitude of E being constant over all or parts of the surface.

5. Calculate the charge enclosed by the gaussian surface. Remember it’s the enclosed charge that matters. Ignore all the charge outside the gaussian surface.

6. Equate the flux to the enclosed charge and solve for E.

FIGURE 22-22 (a) A charged conductor (metal ball) is lowered into an insulated metal can (a good conductor) carrying zero net charge.(b) The charged ball is touched to the can and all of its charge quickly flows to the outer surface of the can.(c) When the ball is then removed, it is found to carry zero net charge.

(A)

Insulator

It4- 41

(c>

22—4 Experimental Basis of Gauss's and Coulomb's Laws

Gauss’s law predicts that any net charge on a conductor must lie only on its surface. But is this true in real life? Let us see how it can be verified experimen­tally. And in confirming this prediction of Gauss’s law, Coulomb’s law is also confirmed since the latter follows from Gauss’s law, as we saw in Section 22-2. Indeed, the earliest observation that charge resides only on the outside of a conductor was recorded by Benjamin Franklin some 30 years before Coulomb stated his law.

A simple experiment is illustrated in Fig. 22-22. A metal can with a small opening at the top rests on an insulator. The can, a conductor, is initially uncharged (Fig. 22-22a). A charged metal ball (also a conductor) is lowered by an insulating thread into the can, and is allowed to touch the can (Fig. 22-22b). The ball and can now form a single conductor. Gauss’s law, as discussed above, predicts that all the charge will flow to the outer surface of the can. (The flow of charge in such situations does not occur instantaneously, but the time involved is usually negligible). These predictions are confirmed in experiments by (1) connecting an electroscope to the can, which will show that the can is charged, and (2) connecting an electroscope to the ball after it has been withdrawn from the can (Fig. 22-22c), which will show that the ball carries zero charge.

The precision with which Coulomb’s and Gauss’s laws hold can be stated quantitatively by writing Coulomb’s law as

F = k Q 1 Q 2J2+ 8

For a perfect inverse-square law, 5 = 0. The most recent and precise experiments (191V) sive 8 = (2.1 + 3.1) X 10-16. Thus Coulomb’s and Gauss’s laws are found

SummaryThe electric flux passing through a flat area A for a uniform electric field E is

<PE = E - A . (22-lb)

If the field is not uniform, the flux is determined from the integral

Gauss’s law states that the net flux passing through any closed surface is equal to the net charge Qenci enclosed by the surface divided by e0:

Q e nclE • dA =eo

(22-4)

= E • dA. (22- 2)

The direction of the vector A or dA is chosen to be perpendic­ular to the surface whose area is A or dA, and points outward from an enclosed surface. The flux through a surface is propor­tional to the number of field lines passing through it.

Gauss’s law can in principle be used to determine the electric field due to a given charge distribution, but its usefulness is mainly limited to a small number of cases, usually where the charge distribution displays much symmetry. The real importance of Gauss’s law is that it is a more general and elegant statement (than Coulomb’s law) for the relation between electric charge and electric field. It is one of the basic equations of electromagnetism.

Questions1. If the electric flux through a closed surface is zero, is the

electric field necessarily zero at all points on the surface? Explain. What about the converse: If E = 0 at all points on the surface is the flux through the surface zero?

2. Is the electric field E in Gauss’s law, <|>E • dA = Qencl/eo> created only by the charge Qenci ?

3. A point charge is surrounded by a spherical gaussian surface of radius r. If the sphere is replaced by a cube of side r, will be larger, smaller, or the same? Explain.

4. What can you say about the flux through a closed surface that encloses an electric dipole?

5. The electric field E is zero at all points on a closed surface; is there necessarily no net charge within the surface? If a surface encloses zero net charge, is the electric field necessarily zero at all points on the surface?

6. Define gravitational flux in analogy to electric flux. Are there “sources” and “sinks” for the gravitational field as there are for the electric field? Discuss.

7. Would Gauss’s law be helpful in determining the electric field due to an electric dipole?

8. A spherical basketball (a nonconductor) is given a charge Q distributed uniformly over its surface. What can you say about the electric field inside the ball? A person now steps on the ball, collapsing it, and forcing most of the air out without altering the charge. What can you say about the field inside now?

9. In Example 22-6, it may seem that the electric field calcu­lated is due only to the charge on the wire that is enclosed by the cylinder chosen as our gaussian surface. In fact, the entire charge along the whole length of the wire contributes to the field. Explain how the charge outside the cylindrical gaussian surface of Fig. 22-15 contributes to E at the gaussian surface. [Hint: Compare to what the field would be due to a short wire.l

10. Suppose the line of charge in Example 22-6 extended only a short way beyond the ends of the cylinder shown in Fig. 22-15. How would the result of Example 22-6 be altered?

11. A point charge Q is surrounded by a spherical surface of radius rQ, whose center is at C. Later, the charge is moved to the right a distance | r 0, but the sphere remains where it was, Fig. 22-23. How is the electricflux OE through the sphere changed? Is the electric field at the surface of the sphere changed? For each “yes” answer, describe the change.

FIGURE 22-23Question 11.

12.

FIGURE 22-24Question 12.

13. A point charge q is placed at the center of the cavity of a thin metal shell which is neutral. Will a charge Q placed outside the shell feel an electric force? Explain.

14. A small charged ball is inserted into a balloon. The balloon is then blown up slowly. Describe how the flux through the balloon’s surface changes as the balloon is blown up. Consider both the total flux and the flux per unit surface area of the balloon.

| Problems22-1 Electric Flux1. (I) A uniform electric field of magnitude 5.8 X 102N /C passes

through a circle of radius 13 cm. What is the electric flux through the circle when its face is (a) perpendicular to thefip.lH linp.s (h \ at 4S° to thp. fip.lH linp.s and ( r \ narallp.l to

2. (I) The Earth possesses an electric field of (average) magni­tude 150 N /C near its surface. The field points radially inward. Calculate the net electric flux outward through asnhp.rir.al snrfar.p snrroimHina and inst hp.vonH thp F.arth’s

A solid conductor carries a net positive charge Q. There is a hollow cavity within the conductor, at whose center is a negative point charge q

—q (Fig. 22-24). What is the charge on /(a) the outer surface of the conductor and (b) the inner surface of the / \conductor’s cavity? [ ( • ] ]

3. (II) A cube of side £ is placed in a uniform field Eq with edges parallel to the field lines, (a) What is the net flux through the cube? (b) What is the flux through each of its six faces?

4. (II) A uniform field E is parallel to the axis of a hollow hemisphere of radius r, Fig. 22-25. (a) What is the electric flux through the hemisphericalsurface? (b) What is the result if E is instead perpendicular to the axis?

AxisI

FIGURE 22-25Problem 4.

22-2 Gauss's Law5. (I) The total electric flux from a cubical box 28.0 cm on a

side is 1.84 X 103N •m2/C. What charge is enclosed by the box?

6. (I) Figure 22-26 shows five closed surfaces that surround various charges in a plane, as indicated. Determine the electric flux through each surface,51,52,53,54, and S$. The surfaces are flat “pillbox” surfaces that extend only slightly above and below the plane in which the charges lie.

FIGURE 22-26Problem 6.

7. (II) In Fig. 22-27, two objects, and 0 2, have charges +1.0 /jlC and -2 .0 /jlC respectively, and a third object, 0 3, is electrically neutral, (a) What is the electric flux through the surface A 1 that encloses all the three objects? (b) What is the electric flux through the surface A 2 that encloses the third object only?

At

FIGURE 22-27Problem 7.

( > , • + 1 .0 /aC

(V -2 ,0 f iC

*n 3Ay

8. (II) A ring of charge with uniform charge density is completely enclosed in a hollow donut shape. An exact copy of the ring is completely enclosed in a hollow sphere. What is the ratio of the flux out of the donut shape to that out of the sphere?

9. (II) In a certain region of space, the electric field is constant in direction (say horizontal, in the x direction), but its magnitude decreases from E = 560 N /C at x = 0 to E = 410 N /C at x = 25 m. Determine the charge within a cubical box of side £ = 25 m, where the box is oriented so that four of its sides are parallelto the field lines (Fig. 22-28). jc= 0 x = 25 m

EFIGURE 22-28Problem 9. -25 tir

10. A noint charge O is nlaced at the center nf a enhe o f

11. (II) A 15.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The total electric flux leaving the bag is 7.3 X 105N*m2/C. What is the linear charge density on the rod?

22-3 Applications of Gauss's Law12. (I) Draw the electric field lines around a negatively charged

metal egg.13. (I) The field just outside a 3.50-cm-radius metal ball is

6.25 X 102N /C and points toward the ball. What charge resides on the ball?

14. (I) Starting from the result of Example 22-3, show that the electric field just outside a uniformly charged spherical conductor is E = a / e 0, consistent with Example 22-8.

15. (I) A long thin wire, hundreds of meters long, carries a uniformly distributed charge of — 7.2 juC per meter of length. Estimate the magnitude and direction of the electric field at points (a) 5.0 m and (b) 1.5 m perpendicular from the center of the wire.

16. (I) A metal globe has 1.50 mC of charge put on it at the north pole. Then —3.00 mC of charge is applied to the south pole. Draw the field lines for this system after it has come to equilibrium.

17. (II) A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of -5 .0 C/m 3. The outer layer has a uniform charge density of +8.0 C/m 3 and extends from an inner radius of6.0 cm to an outer radius of 12.0 cm. Determine the electric field for (a) 0 < r < 6.0 cm, (b) 6.0 cm < r < 12.0 cm, and(c) 12.0 cm < r < 50.0 cm. (d) Plot the magnitude of the electric field for 0 < r < 50.0 cm. Is the field continuous at the edges of the layers?

18. (II) A solid metal sphere of radius 3.00 m carries a total charge of -5.50 fxC. What is the magnitude of the electric field at a distance from the sphere’s center of (a) 0.250 m,(b) 2.90 m, (c) 3.10 m, and (d) 8.00 m? How would the answers differ if the sphere were (e) a thin shell, or ( /) a solid nonconductor uniformly charged throughout?

19. (II) A 15.0-cm-diameter nonconducting sphere carries a total charge of 2.25 /aC distributed uniformly throughout its volume. Graph the electric field E as a function of the distance r from the center of the sphere from r = 0 to r = 30.0 cm.

20. (II) A flat square sheet of thin aluminum foil, 25 cm on a side, carries a uniformly distributed 275 nC charge. What, approximately, is the electric field (a) 1.0 cm above the center of the sheet and (b) 15 m above the center of the sheet?

21. (II) A spherical cavity of radius 4.50 cm is at the center of a metal sphere of radius 18.0 cm. A point charge Q = 5.50 /jlC rests at the very center of the cavity, whereas the metal conductor carries no net charge. Determine the electric field at a point (a) 3.00 cm from the center of the cavity,(b) 6.00 cm from the center of the cavity, (c) 30.0 cm from the center.

22. (II) A point charge Q rests at the center of an uncharged thin spherical conducting shell. What is the electric field E as a function of r (a) for r less than the radius of the shell,(b) inside the shell, and (c) beyond the shell? (d) Does theshell affer.t the field due to (1 alone? D o e s the r.harae O

23. (II) A solid metal cube has a spherical cavity at its center as shown in Fig. 22-29. At the center of the cavity there is a point charge Q = +8.00 /jlC. The metal cube carries a net charge q = -6.10 /xC (not includingQ). Determine (a) the total charge on the surface of the spherical cavity and(b) the total charge on the outer surface of the cube.

FIGURE 22-29Problem 23.

24. (II) Two large, flat metal plates are separated by a distance that is very small compared to their height and width. The conductors are given equal but opposite uniform surface charge densities + cr. Ignore edge effects and use Gauss’s law to show (a) that for points far from the edges, the elec­tric field between the plates is E = cr/e0 and(b) that outside the plates on either side the field is zero, (c) How would your results be altered if the two plates were nonconductors?(See Fig. 22-30).

FIGURE 22-30Problems 24,25, and 26.

25. (II) Suppose the two conducting plates in Problem 24 have the same sign and magnitude of charge. What then will be the electric field (a) between them and (b) outside them on either side? (c) What if the plates are nonconducting?

26. (II) The electric field between two square metal plates is 160 N/C. The plates are 1.0 m on a side and are separated by 3.0 cm, as in Fig. 22-30. What is the charge on each plate? Neglect edge effects.

27. (II) Two thin concentric spherical shells of radii and r2 (t*i < r2) contain uniform surface charge densities ct\ and cr2, respectively (see Fig. 22-31). Determine the electric field for (a) 0 < r < r \ , (b) < r < r2, and(c) r > r2. (<d) Under what conditions will E = 0 for r > r2l (e) Under what conditions will E = 0 for r\ < r < r2l v Neglect the thickness of the shells.

FIGURE 22-31 Two spherical \ shells (Problem 27).

28. (II) A spherical rubber balloon carries a total charge Q uniformly distributed on its surface. At t = 0 the noncon­ducting balloon has radius rQ and the balloon is then slowly blown up so that r increases linearly to 2r0 in a time t. Determine the electric field as a function of time (a) just outside the balloon surface and (b) at r = 3.2r0.

29. (II) Suppose the nonconducting sphere of Example 22-4 has a spherical cavity of radius centered at the sphere’s center (Fig. 22-32). Assuming the charge Q is distributeduniformly in the “shell” (between r = r\ and r = r0), determine the electric field as a function of r for (a) 0 < r < rx, (b) rx < r < r0, and (c) r > r0.

FIGURE 22-32Problems 29,30,31, and 44.

30. (II) Suppose in Fig. 22-32, Problem 29, there is also a charge n at the cen ter nf the cavitv. D eterm in e the electric

31. (II) Suppose the thick spherical shell of Problem 29 is a conductor. It carries a total net charge Q and at its center there is a point charge q. What total charge is found on(a) the inner surface of the shell and (b) the outer surface of the shell? Determine the electric field for (c) 0 < r < rl5(id) rx < r < r0, and (e) r > r0.

32. (II) Suppose that at the center of the cavity inside the shell (charge Q) of Fig. 22-11 (and Example 22-3), there is a point charge q ±Q). Determine the electric field for(a) 0 < r < r0, and for (b) r > r0. What are your answers if (c) q = Q and (d) q = - Q l

33. (II) A long cylindrical shell of radius R q and length £ (Rq « . £) possesses a uniform surface charge density (charge per unit area) cr (Fig. 22-33). Determine the electric field at points (a) outside the cylinder (R > Rq) and (b) inside the cylinder (0 < R < R0); assume thepoints are far from the ends and not , + + ^ ^too far from the shell ( / ? « £ ) . + + + +(c) Compare to the result for a long + + + +line of charge, Example 22-6. Neglect | + + + +the thickness of shell. + + + +

FIGURE 22-33 + + + +Problem 33. + + + +

34. (II) A very long solid nonconducting cylinder of radius R0 and length £ (R0 « £) possesses a uniform volume charge density pE (C/m3), Fig. 22-34. Determine the electric field at points (a) outside the cylinder (R > R0) and (b) inside the cylinder (R < R0). Do only for points far from the ends and for which R « L

FIGURE 22-34Problem 34.

/ + \ + + + + + +M oU>+ + + + + + +

\ + / + + + + + +

35. (II) A thin cylindrical shell of radius R\ is surrounded by a second concentric cylindrical shell of radius R2 (Fig. 22-35). The inner shell has a total charge + Q and the outer shell —Q. Assuming the length £ of the shells is much greater than R\ or R2, determine the electric field as a function of R (the perpendicular distance from the common axis of the cylin­ders) for (a) 0 < R < R i , (b) Ri < R < R2, and (c) R > R2.(d) What is the kinetic energy of anelectron if it moves between (andconcentric with) the shells in a circularorbit of radius (Ri + R2)/21 Neglect / r ^thickness of shells. \

r/i<iFIGURE 22-35Problems 35,36, and 37.

36. (II) A thin cylindrical shell of radius Ri = 6.5 cm is surrounded by a second cylindrical shell of radius R2 = 9.0 cm, as in Fig. 22-35. Both cylinders are 5.0 m long and the inner one carries a total charge Q\ = —0.88 /jlC and the outer one Q2 = +1.56 fxC. For points far from the ends of the cylinders, determine the electric field at a radial distance r from the central axis of (a) 3.0 cm, (b) 7.0 cm, and (c) 12.0 cm.

37. (II) (a) If an electron (m = 9.1 X 10-31kg) escaped from the surface of the inner cylinder in Problem 36 (Fig. 22-35) with negligible speed, what would be its speed when it reached the outer cylinder? (b) If a proton (m = 1.67 X 10-27kg)revolves in a circular nrhit o f radius r = 7 0 cm about the

38. (II) A very long solid nonconducting cylinder of radius Rx is uniformly charged with a charge density pE. It is surrounded by a concentric cylindrical tube of inner radius R2 and outer radius R3 as shown in Fig. 22-36, and it too carries a uniform charge density pE. Determine the electric field as a function of the distance R from the center of the cylinders for (a) 0 < R < R lt (b) R x < R < R2,(c) R2 < R < R3, and (d) R > R3. (e) If pE = 15pC /m 3 and Ri = \ R 2 = \ R 3 = 5.0 cm, plot £ as a function of R from R = 0 to R = 20.0 cm. Assume the cylinders are very long compared to R3.

FIGURE 22-36Problem 38.

39. (II) A nonconducting sphere of radius r0 is uniformly charged with volume charge density pE. It is surrounded by a concentric metal (conducting) spherical shell of inner radius rx and outer radius r2, which carries a net charge +Q. Determine the resulting electric field in the regions(a) 0 < r < r0, (b) r0 < r < rx, (c) rx < r < r2, and(d) r > r2 where the radial distance r is measured from the center of the nonconducting sphere.

40. (II) A very long solid nonconducting cylinder of radius Rx is uniformly charged with charge density pE. It is surrounded by a cylindrical metal (conducting) tube of inner radius R2 and outer radius R3, which has no net charge (cross-sectional view shown in Fig. 22-37). If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resultingelectric field in the region R > R3, where the radial distance R is measured from the metal cylinder’s axis.Assume d < (R2 - i^).

FIGURE 22-37Problem 40.

41. (II) A flat ring (inner radius R0, outer radius 4R0) is uniformly charged. In terms of the total charge Q, deter­mine the electric field on the axis at points (a) 0.25R0 and(b) 75Rq from the center of the ring. [Hint: The ring can be replaced with two oppositely charged superposed disks.]

42. (II) An uncharged solid conducting sphere of radius r0 contains two spherical cavities of radii rx and r2, respec­tively. Point charge Q\ is then placed within the cavity of radius rx and point charge Q2 is placed within the cavity of radius r2 (Fig.22-38). Determine the resulting electric field (magnitude and direc­tion) at locations outside the solid sphere (r > r0), . where r is the distance from its center.

FIGURE 22-38

43. (Ill) A very large (i.e., assume infinite) flat slab of noncon­ducting material has thickness d and a uniform volume charge density +pE. (a) Show that a uniform electric field exists outside of this slab. Determine its magnitude E and its direc­tion (relative to the slab’s surface), (b) As shown in Fig. 22-39, the slab is now aligned so that one of its surfaces lies on the line y = x. At time t = 0, a pointlike particle (mass ra, charge +q) is located at position r = + y0j and has velocity v = v0i. Show that the particle will collide with the slab if v0 ^ V V 2 q y 0pEd/e0 ra.Ignore gravity.

FIGURE 22-39 +Problem 43.

44. (Ill) Suppose the density of charge between rx and r0 of the hollow sphere of Problem 29 (Fig. 22-32) varies as Pe = Pori / r • Determine the electric field as a function of r for (a) 0 < r < rx, (b) rx < r < r0, and (c) r > r0. (d) Plot E versus r from r = 0 to r = 2r0.

45. (Ill) Suppose two thin flat plates measure 1.0 m X 1.0 m and are separated by 5.0 mm. They are oppositely charged with + 15 fiC. (a) Estimate the total force exerted by one plate on the other (ignore edge effects), (b) How much work would be required to move the plates from 5.0 mm apart to1.00 cm apart?

46. (Ill) A flat slab of nonconducting material (Fig. 22-40) carries a uniform charge per unit volume, pE. The slab has thickness d which is small compared to the v height and breadth of the slab. Determine the electric field as a function of x (a) inside the slab and (b) outside the slab (at distances much less than the slab’s height or breadth).Take the origin at the center of the slab.

FIGURE 22-40Problem 46.

-( / •

47. (Ill) A flat slab of nonconducting material has thickness 2d, which is small compared to its height and breadth. Define the x axis to be along the direction of the slab’s thickness with the origin at the center of the slab (Fig. 22-41). If the slab carries a volume charge density pE(x) = — p0 in the region —d < x < 0 and pE(jc) = +p0 in the region0 < x < +d, determine the electric field E as a function of x in the regions (a) outside the slab, (b) 0 < x < +d, and(c) —d < x < 0. Let po be a positive constant.

FIGURE 72-A1

V

+ +

“ - + +

- -

- - + +

- - + +

- d ■ . 0 + + X

- - + +

- - + +

- -

48. (Ill) An extremely long, solid nonconducting cylinder has a radius R0. The charge density within the cylinder is a function of the distance R from the axis, given by pE (R) = p0(R /R0)2. What is the electric field everywhere inside and outside the cylinder (far away from the ends) in terms of p0 and R07

49. (Ill) Charge is distributed within a solid sphere of radius r0 in such a way that the charge density is a function of the radial position within the sphere of the form: pE(r) = po(r/r0). If the total charge within the sphere is Q (and positive), what is the electric field everywhere within the sphere in terms of Q, r0, and the radial position rl

| General Problems50. A point charge Q is on the axis of a short cylinder at its

center. The diameter of the cylinder is equal to its length i (Fig. 22-42). What is the total flux through the curved sides of the cylinder? [Hint. First calculate the flux through the ends.] t — 2/?0

58. Three large but thin charged sheets are parallel to each other as shown in Fig. 22-44. Sheet I has a total surface charge density of 6.5nC/m2, sheet II a charge of —2.0nC/m2, and sheet III a charge of 5.0nC/m2. Estimate the force per unit area on each sheet, in N/m2?

QFIGURE 22-42Problem 50.

51. Write Gauss’s law for the gravitational field g (see Section 6-6).

52. The Earth is surrounded by an electric field, pointing inward at every point, of magnitude E « 150 N/C near the surface, (a) What is the net charge on the Earth? (b) How many excess electrons per square meter on the Earth’s surface does this correspond to?

53. A cube of side i has one corner at the origin of coordinates, and extends along the positive x, y, and z axes. Suppose the electric field in this region is given by E = (ay + b)\. Determine the charge inside the cube.

54. A solid nonconducting sphere of radius rQ has a total charge Q which is distributed according to pE = br, where pE is the charge per unit volume, or charge density (C/m3), and b is a constant. Determine (a) b in terms of Q, (b) the electric field at points inside the sphere, and (c) the electric field at points outside the sphere.

55. A point charge of 9.20 nC is located at the origin and a second charge of — 5.00 nC is located on the x axis at x = 2.75 cm. Calculate the electric flux through a sphere centered at the origin with radius 1.00 m. Repeat the calcu­lation for a sphere of radius 2.00 m.

56. A point charge produces an electric flux of +235N-m2/C through a gaussian sphere of radius 15.0 cm centered on the charge, (a) What is the flux through a gaussian sphere with a radius 27.5 cm? (b) What is the magnitude and sign of the charge?

57. A point charge Q is placed a distance r0/2 above the surface of an imaginary spherical surface of radius r0 (Fig. 22-43).(a) What is the electric flux through the sphere? (b) What range of values does E have at the surface of the sphere?(c) Is E perpendicular to the sphere at all points? (d) Is Gauss’s law useful for obtaining E at thesurface of the sphere? j ̂ rt,

FIGURE 22-44Problem 58.

IIIm

59. Neutral hydrogen can be modeled as a positive point charge +1.6 X 10-19C surrounded by a distribution of negative charge with volume density given by p^(r) = —A e~ 2r^a° where a0 = 0.53 X 10-10 m is called the Bohr radius, A is a constant such that the total amount of negative charge is —1.6 X 10-19 C, and e = 2.718 ••• is the base of the natural log. (a) What is the net charge inside a sphere of radius aQ ?(b) What is the strength of the electric field at a distance a0 from the nucleus? [Hint: Do not confuse the exponential number e with the elementary charge e which uses the same symbol but has a completely different meaning and value (e = 1.6 X 10“19C).]

60. A very large thin plane has uniform surface charge density a. Touching it on the right (Fig. 22-45) is a long wide slab of thickness d with uniform volumecharge density pE. Determine the electric field(a) to the left of the plane, (b) to the right of the slab, and (c) everywhere inside the slab.

FIGURE 22-45Problem 60.

P E

- i t -

61. A sphere of radius r0 carries a volume charge density pE (Fig. 22-46). A spherical cavity of radius r0/2 is then scooped out and left empty, as shown, (a) What is the magnitude and direction of the electric field at point A?(b) What is the direction and magnitude of the electric field at point B? Points A and C are at the centers of the respective spheres. / pE

R< ♦A -C

FIGURE 22-46Problem 61.

FIGURE 2 2 -4 3

62. Dry air will break down and generate a spark if the electric field exceeds about 3 X 106N/C. How much charge couldhe narked onto the surf a re o f a oreen nea ^diameter

63. Three very large sheets are separated by equal distances of 15.0 cm (Fig. 22-47). The first and third sheets are very thin and nonconducting and have charge per unit area a of +5.00 f j i C / m 2 and -5.00 /iC /m 2 respectively. The middle sheet is conducting but has no net charge. (a) What is the electric field inside the middle sheet? What is the electric field (b) between the left and middle sheets, and (c) between the middle and right sheets? ((d) What is the charge density on the surface of the middle sheet facing the left sheet, and (e) on the surface facing the right sheet?

- +5.00 fiCJm2 {Jnel = 0 - -5-00 /*C/m2 * I- +1 W— 15.0 cm— — 15.0 cm —

+ — —

FIGURE 22-47 Problem 63.

64. In a cubical volume, 0.70 m on a side, the electric field is

E = £o( l + f ) i + £og ) j

where E0 = 0.125 N /C and a = 0.70 m. The cube has its sides parallel to the coordinate axes, Fig. 22-48. Determine the net charge within the cube.

FIGURE 22-48 /Problem 64.

65. A conducting spherical shell (Fig. 22-49) has inner radius = 10.0 cm, outer radius = 15.0 cm, and has a +3.0 fxC point charge at the . - —center. A charge of -3 .0 jjlC is put on the conductor, (a) Where on the conductor does the -3 .0 fj,C end up? (b) What is the electric field both inside and outside the shell?

FIGURE 22-49Problem 65.

/ArJCUO cm \ \

C? 15.0)cm7^\

y

66. A hemisphere of radius R is placed in a charge-free region of space where a uniform electric field exists of magnitude E directed perpendicular to the hemisphere’s circular base (Fig. 22-50). (a) Using the definition of through an “open” surface, calculate (via explicit integration) the electric flux through the hemisphere. [Hint: In Fig. 22-50 you can see that, on the surface of a sphere, the infinitesimal area located between the angles 0 and 0 + dd is dA = (2ttR sin 6)(Rdd) = 2ttR2 sin 6 d0.\ (b) Choose an appropriate gaussian surface and use Gauss’s law to much more easily obtain the same result for the electric flux through the hemisphere.

FIGURE 22-50Problem 66.

* Numerical/Computer* 67. (Ill) An electric field is given by

(x+y)2 (x+yVE = E x o e \ a J i + Ey0e \ a ) J’

where Exq = 50 N/C, Eyo = 25 N/C, and a = 1.0 m. Given a cube with sides parallel to the coordinate axes, with one corner at the origin (as in Fig. 22-48), and with sides of length 1.0 m, estimate the flux out of the cube using a spreadsheet or other numerical method. How much total charge is enclosed by the cube?

Answers to Exercises

A: (d). D: (e).B: («). E: (c).