Chapter 27 by John Smith

GAUSS'SLAW

27.1. V i a l i e :

As discussed in Section 27.1, the symmetry of the electric field must match the symmetry of the charge distribution. In particular, the electric field of a cylindrically symmetric charge distribution cannot have a component parallel to the cylinder axis. Also, the electric field of a cylindrically symmetric charge distribution cannot have a component tangent to the circular cross section. The only shape for the electric field that matches the symmetry of the charge distribution with respect to (i) translation parallel to the cylinder axis, (ii) rotation by an angle about the cylinder axis, and (iii) reflections in any plane containing or perpendicular to the cylinder axis is the one shown in the figure.

27.2. Visualize:

+++++++++++++++++ +++++++++++++++++ 1!11!1111111111

Figure 27.6 shows the electric field for an infinite plane of charge. For two parallel planes, this is the only shape of the electric field vectors that matches the symmetry of the charge distribution.

27.4. Model: The electric flux "flows" our of a closed surface around a region of space containing a net positive charge and inro a closed surface surrounding a net negative charge. Visualize: Please refer to Figure Ex27.4. Let A be the area in m2 of each of the six faces of the cube. Solve: The electric flux is defined as = = EAcose, where 8 is the angle between the electric field and a line perpendiculur to the plane of the surface. The electric flux our of the closed cube surface is (Pa",= (20 N / C + 20 N / C + 10 N /C)AcosO" = (50A) N m' / C Similarly. the electric tlux inro the closed cube surface is

â&#x201A;Ź.A

= (15 N / C

+ 15 N / C + 15 N / C)Acos180Â° = - ( U A )

N m' / C

The net electric flux is ( 5 0 A ) N m' / C - (45A) N m' / C = (5A) N m' / C . Since the net electric tlux is positive (i.e.. outward). the closed box contains a positive charge.

27-1

27-2

Chapter 27

27.5. Model: The electric flux "flows" out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure Ex27.5. Let A be the area of each of the six faces of the cube. Solve: The electric flux is defined as Qe = E . 2 = EAcose, where 8 is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is Qoul

= (10 N / C +10 N / C + l O N / C + 5 N / C)AcosO" = (35A) N m2 / C

Similarly, the electric flux into the closed cube surface is Qin=(15 N/C+20N/C)Acos(l8O0)=-(35A)Nm2 / C

Hence,

Qoul

Qin

= 0 N m2/C. Since the net electric flux is zero, the closed box contains no charge.

27.6. Model: The electric flux "flows" out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure Ex27.6. Let A be the area of each of the six faces of the cube. Solve: The electric flux is defined as Qe = ~?.i = EAcose, where 8 i s the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is Qout

= (20 N / C + 10 N / C + 1 0 N / C)AcosO" = (40A) N m2 / C

Similarly, the electric flux into the closed cube surface is Q,,, = (20 N / C + 15 N / C)Acos180째 = -(35A) N m2 / C

+ Qm

Because the cube contains negative charge, Qom Therefore,

must be negative. This means

(40A) N m2 / C+(-35A) N m2 / C+Qbo,,

* Qbow

Qom

+ Q," +

< 0 N m2/C.

< 0 N m2 / C

< (-5A) N m2 / C

That is, the unknown vector points into the front face of the cube and its minimum possible field strength is 5 N/C.

27.7. Model: The electric flux "flows" out of a closed surface around a region of space containing a net positive charge and into a closed surface surrounding a net negative charge. Visualize: Please refer to Figure Ex27.7. Let A be the area of each of the six faces of the cube. Solve: The electric flux is defined as Qe = E.2 = EAcos8, where 8 is the angle between the electric field and a line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is =(15N/C + 1 5 N / C + 10N/C)Acos0째=(40A)Nm2/C Similarly, the electric flux into the closed cube surface is QoUl

a,,, = (20 N / C +

15 N / C)Acos180째 = -(35A) N m2 / C

Because the cube contains no charge, the total flux is zero. This means Qln

+ QouI +

= 0 N m'/C

QbOw

= -(5A) N m2 / C

The unknown electric field vector on the front face points in, and the electric field strength is 5 N/C.

27.8. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure Ex27.8. The electric field vectors make an angle of 30" with the planar surface. to the planar surface is at an angle of 90" with the surface, the angle between 6 and E is Because the normal e = 600. Solve: The electric flux is Qe

. i = E A c o s 8 = ( 2 0 0 N / C ) ( 1 . 0 ~ 1 0 -m2)cos60"=1.0Nm2 ~ /C

27.9. Model: The electri'c field is uniform over the entire surface. Visualize: Please refer to Figure Ex27.9. The electric field vectors make an angle of 30" below the surface. Because the normal to the planar surface is at an angle of 90" relative to the surface, the angle between Solve: The electric flux is

oe= E - i= m c o s e = (200 N

cl(1.0 x IO-? m z ) c o s 1 2 ~=o -1.0 N m' / c

and

E is 8= 120".

Gauss's Law

27-3

27.10. Model: The electric field is uniform over the entire surface. Visualize: Please refer to Figure 27.10. The electric field vectors make an angle of 60" above the surface. Because the normal 2 to the planar surface is at an angle of 90" relative to the surface, the angle between 6 and ?i is O= 30". Solve: The electric flux is @e

=E.~=EAcosO~15.0Nm / C2 = E(l.OxlO-' m 2 ) c o s 3 0 " ~ E = 1 7 3 0 N / C

27.11. Model: The electric field is uniform over the rectangle in the xy plane. Solve: (a) The area vector is perpendicular to the xy plane. Thus 2 = (2.0 cm x 3.0 cm)i = (6.0 x lo4 m 2 ) i The electric flux through the rectangle is

De = E.2 = (50;

lOOi) .(6.0 x l0"i) N m2 / C = 6.0 x lo-' N m' / C

(b) The electric flux is

me = E . 2 = ( 5 0 1 + Assess: In (b),

100~)~(6.0x10'k N m ' / C = O N m ' / C

E is in the plane of the rectangle. That is why the flux is zero.

27.12. Model: The electric field over the rectangle in the xz plane is uniform. Solve: (a) The area vector is perpendicular to the xz plane. Thus 2 = (2.0 cm x 3.0 c m ) j = (6.0 x lo4 m 2 ) j The electric flux through the rectangle is @e

=,??-2=(501+ 100i)N/C~(6.Ox1O4m 2 ) j =(3O0x1O4 N r n 2 / C ) ( ~ . ~ ) + ( 6 0 0 x 1 ON4m ' / C ) ( ~ . ~ ) = O N m * / C

(b) The flux is @e

E . 2 = (50; + l 0 0 j ) N / C.(6.0 x lo4

m')j

=(3O0x1O4 N m 2 / C ) ( ~ . ~ ) + ( 6 0 0 x 1 0N4 m 2 / C ) ? . j = 0 N m2 /C(0.06 N m2 / C) = 0.06 N m2 / C

27.13. Model: The electric field over the circle in the xy plane is uniform. Solve: The area vector of the circle is

2 = nr2i = ~ ( 0 . 0 2m ) 2 i =

(0.001257 m 2 ) i

Thus, the flux through the area of the circle is

me = E . i = ( l O O O ~ + l O O O ~ + l O O O ~ ) N / C - ( 0 . 0 0 1 2m')L 57 A

*..

Using i . k = j . k = Oand k . k = l , @e

= (1000 N / C)(0.001257 m') = 1.26 N m' / C

27.14. Model: The electric field over the two faces of the box perpendicular to Visualize:

E is uniform.

27-4

Chapter 27

Solve: The electric field is parallel to four sides of the box, so the electric fluxes through these four surfaces are zero. The field is out of the top surface, so @q = EA. The field is into the bottom surface, so Qbmm = -EA. Thus the net flux is De= 0 N m2/C2.

27.15. Model: The electric field through the two cylinders is uniform. Visualize: Please refer to Figure Ex 27.15 . Let A = nR2 be the area of the end of the cylinder and let E be the electric field strength.

Solve: (a) There's no flux through the side walls of the cylinder because

is parallel to the wall. On the right

end, where E' points outward, = EA = nR2E. The field points inward on the left, so @,er, = -EA = -rR2E. Altogether, the net flux is @e = 0 N m'/C2. --*

= EA = nR2E. Thus the

(b) The only difference from part (a) is that E points outward on the left end, making

net flux through the cylinder is @e = 2nR'E.

27.16. Solve: For any closed surface enclosing a total charge Qm,the net electric flux through the surface is

* Q,

me =-

n I '

= gome = (8.85 x

C2 / Nm2)(-1000 N m 2 / C) = -8.85 nC

27.17. Solve: For any closed surface that encloses a total charge Qh, the net electric flux through the surface is @ e

(55.3 X 106)(-1.60 X lo-'' c) Q. =a= =-1.0Nm2/C â&#x201A;Ź0 8.85 x lo-" C2 / N m 2

27.18. Solve: For any closed surface that encloses a total charge Q,,, the net electric flux through the closed

me = Q,,,/E, . The flux through the top surface of the cube is one-sixth of the total: 10 x io-' c - Qin = 188 N m */ C

surface is

@ e slllface

6 ~ , 6(8.85 x lo-'* C2 / N m 2 )

27.19. Solve: For any closed surface that encloses a total charge Qin,the net electric flux through the closed surface is @e

me = ',,,/E,

. The total flux through the cube is = (600 N m2 / C)(8.85 x lo-'' C2 / N m 2 ) = 5.31 x

= 6(100 N m2

~ o c- =~ 5.31 nC

27.20. Visualize: I

0 4

:,_o-

oq (Cf

I , ,

For any closed surface that encloses a total charge Qin, the net electric flux through the closed surface is @e = Q,,/Eo .

Gauss’s Law

27-5

27.21. Visualize:

@ 2q ,

‘

0 - 4

‘i

‘_-_-’

For any closed surface that encloses a total charge Q,, the net electric flux through the closed surface is Q e = Qm / E o

27.22. Visualize: Please refer to Figure Ex27.22. For any closed surface that encloses a total charge Qln,the net electric flux through the closed surface is Qe = Qin/&,. For the closed surface of the torus, Q, includes only the -1 nC charge. So, the net flux through the torus is due to this charge: Qe =

- 1 ~ 1 0 -c~ = -113 N m2 / C 8.85 x lo-” C2 / Nm2

This is inward flux.

27.23. Visualize: Please refer to Figure Ex27.23. For any closed surface that encloses a total charge Q,,, the net

. The cylinder encloses the +1 nC charge only as both the electric flux through the closed surface is @ e = +lo0 nC and the -100 nC charges are outside the cylinder. Thus, Qe =

1 . 0 ~ 1 0 -c~ = 113 N m2 / C . 8.85 x lo-’’ Cz / Nm2

This is outward flux.

27.24. Model: A copper penny is a conductor. The excess charge on a conductor resides on the surface. Solve: The electric field at the surface of a charged conductor is

q = E ~ E =~(8.85 , ~x lo-’’ ~ ~ C 2 / Nm2)(2000N / C) = 17.7 x

C / m2

27.25. Model: The excess charge on a conductor resides on the outer surface. Solve: The electric field at the surface of a charged conductor is perpendicular to surface

q = E,E,,,,

= (8.85 x lo-” C’ / Nm2)(3x lo6 N / C) = 2.66 x lo-’ C / m’

Assess: It is the air molecules just above the surface that “break down” when the E-field becomes strong enough to accelerate stray charges to approximately 15 eV between collisions, thus causing collisional ionization. It does not make any difference whether E points toward or away from the surface.

27.26. Model: The excess charge on a conductor resides on the outer surface. Visualize: Please refer to Figure P27.26. Solve: Point 1 is at the surface of a charged conductor, hence perpendicular to surface

(5.0 x 101’)(1.60x

C/m‘)

8.85 x IO-’’ C’ / N m’

= 904 N /

At point 2 the electric field strength is zero because this point lies inside the conductor. The electric field strength at point 3 is zero because there is no excess charge on the interior surface of the box. This can be quickly seen by considering a Gaussian surface just inside the interior surface of the box as shown in Figure 27.3 1.

“-._I.^

. .... .,

”-

27-6

Chapter 27

27.27. Model: The electric field over the four faces of the cube is uniform. Visualize: Please refer to Figure â&#x201A;Ź27.27. = EAcos8, where 8 is the angle between the Solve: (a) The electric flux through a surface area 2 is Qe = i?.i electric field and the vector ;? that points outward from the surface. Thus Q, = EAcos8, = (500 N / C)(9.0 x lo" m2)cos150"= -0.390N m2 / C Q2

= EAcos8, = (500 N / C)(9.0 x lo" m2)cos60"= 0.225 N m2 / C

Q3 = EAcos8,

= (500 N / Cl(9.0 x lo4 m2)cos30" = 0.390 N m2 / C

Q, = EAcos8, = (500 N / C)(9.0 x IO4 m2)cos120" = -0.225 N m2 / C (b) The net flux through these four sides is Q e = Q, + Q 2 + Q 3 + Q4 = 0 N m2 / C. Assess: The net flux through the four faces of the cube is zero because there is no enclosed charge.

27.28. Model: The electric field over the five surfaces is uniform. Visualize: Please refer to Figure P27.28. Solve: The electric flux through a surface area 2 is Qe = i.i= EAcos8 where 8 is the angle between the electric field and a line perpendicular to the plane of the surface. Because the electric field is perpendicular to side 1 and is parallel to sides 2, 3, and 5. Also the angle between E and is 60". The electric fluxes through these five surfaces are

0,= E,A, cos8, = (500 N / C)(1 m x 2 m)cos(l80") = -1000 N m2 / C Q2

=E2A,cos900=Q,=Q5=ONm2/C

Q, = E4A4cos8, = (500 N / C)((l m/sin30") x 2 m)cos60" = +lo00 N m2 / C

Assess: Because the flux into these five faces is equal to the flux out of the five faces, the net flux is zero, as we found.

4/+

27.29. Model: Because the tetrahedron contains no charge, the net flux through the tetrahedron is zero. *

Visualize:

-- -+-- 1

60" a Abase =

h = a cos 60" = .-1a5 1 ah. =1 5 a' 2 4

Abase

Solve: (a) The area of the base of the tetrahedron is + a Z , where a = 20 cm is the length of one of the sides. Because the base of the tetrahedron is parallel to the ground and the vertical uniform electric field passes upward through the base, the angle between E and ;? is 180". Thus,

- -

@base

= E.Aba, = EA,,,cos180"=-E--a2

=-(200N/C)-(0.20m)2

=-3.46 N m 2 / C

(b) Since E is perpendicular to the base, the other three sides of the tetrahedron share the flux equally. Because the tetrahedron contains no charge, Qne, = abase + 3Q,,, = 0 N m' / C -3.46 N m2/ C + 3Qsrdr= 0 N m / C -QSlde

=1.15Nm2/C

Gauss's Law

27-7

27.30. Visualize: Please refer to Figure P 27.30. Solve: For any closed surface that encloses a total charge Q,", the net electric flux through the surface is CDe = Q, / E ,

. We can write three equations from the three closed surfaces in the figure:

Subtracting third equation from the first, 41 - 92 = +9

Adding second equation to this equation, 2q, = +4q

q1= 2q

That is, q1= +2q, q2= +q, and q3= -3q.

27.31. Solve: The sphere encloses both q, and q2.For any closed surface that encloses a total charge Q,",the net electric flux through the closed surface is

Q,,, - (-4Q+ 2Q) ---2Q =--

@ e

27.32. Solve: (a) The electric field is

E = (5000 r 2 ) ;N / C = 5000 (0.20)2; N / C = 200; N / C So the electric field strength is 200 N/C. (b) The area of the surface is Asphere= 4z(0.20 m)' = 0.5027 m'. Thus, the electric flux is

- -

CD, =fE.dA=EAsphae =(200N/C)(0.5027m2)=100.5Nm2/C=101N m 2 / C (c) Because CDe =

e,,/&,, , Q,. = qCPe= (8.85 x lo-'' C2/Nm2)(100.5 N m2/C) = 8.90 x lo-'' C

27.33. Solve: (a) The electric field is

So the electric field strength is 2000 N/C. (b) The area of the spherical surface is Asphere= 4z(0.10 m)' = 0.1257 m2. Hence, the flux is CD, = (c) Because

= Q,./

$2. d i = EAsphere = (2000 N / C)(0.1257 m ') = 25 1 N m' / C

E ,~

Q,,= qCDe= (8.85 x lo-'' C'/N m2)(251N mZ/C)= 2.22 x

C = 2.22 nC

27.34. Model: The copper plate is a conductor. The excess charge resides on the surface of the plate. Solve: (a) One-half of the electrons are located on the top surface and one-half on the bottom surface of the copper plate, so the surface charge density is 77=

(0.5x10'0)(1.60x10~'9C)

0.1 m x 0.1 m Thus, the electric field at the surface of the plate is 17 E=-=

= 8 0 ~ 1 0C - ~/ m '

80x10" C / m ' =9040N/C 8.85 x lo-'' C' / N m 2 Because the charge on the plate is negative, the direction of the electric field is toward the plate. (b) E = 0 N/C because the electric field within a conductor is zero. ( c ) The electric field E = 9040 N/C, toward the plate. E,

27-8

Chapter 27

2735. Model: The excess charge on a conductor resides on the outer surface. Visualize: Before After

, 100 nC

Gaussian surface

Solve: (a) Consider a Gaussian surface surrounding the cavity just inside the conductor. The electric field inside a conductor in electrostatic equilibrium is zero, so E' is zero at all points on the Gaussian surface. Thus @e = 0. Gauss's law tells us that @e = e% /, , so the net charge enclosed by this Gaussian surface is Q,, = QWmt + Q,,,, = 0. We know that QWm,= +lo0 nC, so Qwal,= -100 nC. The positive charge in the cavity attracts an equal negative charge to the inside surface. (b) The conductor started out neutral. If there is -100 nC on the wall of the cavity, then the exterior surface of the conductor was initially +IO0 nC. Transferring -50 nC to the conductor reduces the exterior surface charge by 50 nC, leaving it at +50 nC. Assess: The electric field inside the conductor stays zero.

27.36. Visualize: Please refer to Figure P27.36. Solve: For any closed surface that encloses a total charge Qm,the net electric flux through the closed surface is @, = In the present case, the conductor is neutral and there is a point charge Q inside the cavity. Thus Q,=Qandtheflwcis = -Q

@ e

27.37. Model: The ball is uniformly charged. The charge distribution is spherically symmetric. Visualize:

Spherical Gaussian surfaces

The figure shows spherical Gaussian surfaces with radii r = 5 cm, 10 crn, and 20 cm. These surfaces match the symmetry of the charge distribution. So, E is perpendicular to the Gaussian surface and the value of the field strength is the same at all points on the surface. Solve: (a) Because the ball is uniformly charged, its charge density is p = -Q -- V

Q =

Sm-'

3(80 x

= 2.39 x 10" C / m3

4~(0,20m)~

(b) Once again, because the ball is uniformly charged,

("3"

Q = pV = p( $ r 3 ) = (2.387 x 10" C / m3) -

= (1.00 x lo-' C / m')r7

m e n r = 5 c m , Q5= ( l . O 0 ~ 1 0 C/m')(0.05m)3=1.25nC.Whenr=10crnand20cm, -~ Qlo=lO.OnC~Q d ro= 80.0 nC .

GUUSS’S Law

f E . d i = EA = Q, / E ~ For . the 5.0-cm-radius EoAs

Gaussian spherical surface,

1 . 2 5 ~ 1 0 -C~

= EsAs jE5 = Q5 (8.85 x

27-9

= 4500 N / C

IO-” C2 / Nm’)[4~(0.05m)’]

Similarly, we can apply Gauss’s law to the 10.0-cm-radius and 20.0-cm-radius spherical surfaces and obtain E,, = 9,000 N / C and Ez0 = 18,OOO N / C. = 2EI0= 4Es. This result is consistent with the result of Example 27.4 according to which Assess: We note that

27.38. Model: The excess charge on a conductor resides on the outer surface. The field inside, outside, and within the hollow metal sphere has spherical symmetry. Visualize: E

Spherical Gaussian

surfaces

The figure shows spherical Gaussian surfaces with radii r I a, a < r < b, and r 2 b. These surfaces match the symmetry of the charge distribution. Solve: (a) For r 5. u. Gauss’s law is

Notice that the electric field is everywhere perpendicular to the spherical surface. Because of the spherical symmetry of the charge, the electric field magnitude E is the same at all points on the Gaussian surface. Thus,

where we made use of the fact that E is directed radially outward. The field depends only on the enclosed charge, not on the charge on the outer sphere. For (I c r < b, Gauss’s law is

- -

Q,” = f E . dA = EAspherc = E( 4 d ) = EO

Here Q,,= 0 C. It is not +Q, because the charge in the cavity polarizes the metal sphere in such a way that E = 0 in the metal. Thus a charge -Q moves to the inner surface. Because the hollow sphere has a net charge of +2Q, the exterior surface now has a charge of +3Q. Thus, the electric field E = 0 N/C. For r 2 b, Q,, = QcxWrtor + Q,nlenor + QLartl) = +3Q + + = +3Q

(-e) (+e)

Chapter 27

27-10

Gauss's law applied to the Gaussian surface at r 1 b yields: ...Qi,, - +3Q me = E . dA = EAsphere= E ( 4 m 2 )= - -jE =

y);

1 3 Q 1 T )a E- = 4ZEO 4 m 0 E, Eo (b) As determined in part (a), the inside surface of the hollow sphere has a charge of -Q, and the exterior surface of the hollow sphere has a charge of +3Q. Assess: The hollow sphere still has the same charge +2Q as given in the problem, although the sphere is polarized.

27.39. Model: The excess charge on a conductor resides on the outer surface. The charge distribution on the two spheres is assumed to have spherical symmetry. Visualize: Please refer to Figure P27.39. The Gaussian surfaces with radii r = 8 cm, 10 cm, and 17 cm match the symmetry of the charge distribution. So, E is perpendicular to these Gaussian surfaces and the field strength has the same value at all points on the Gaussian surface. Solve: (a) Gauss's law is me = dA = . Applying it to a Gaussian surface of radius 8 cm,

fz. e,,,/&,

Q,,, = - E , E A , ~ ~ ~=, -(8.85 x

C2 / N m2)(15,000N / C)[4~(0.08m)2]= -1.068 x IO-' C

Because the excess charge on a conductor resides on its outer surface and because we have a solid metal sphere inside our Gaussian surface, Qinis the charge that is located on the exterior surface of the inner sphere. (b) In electrostatics, the electric field within a conductor is zero. Applying Gauss's law to a Gaussian surface just inside the inside-surface of the hollow sphere at r = 10 cm,

That is, there is no net charge. Because the inner sphere has a charge of -1.068 x lo4 C, the inside surface of the hollow sphere must have a charge of +1.068 x IO4 C. (c) Applying Gauss's law to a Gaussian surface at r = 17 cm, Q, = E ~ $ EdA. = E,EA,,, = (8.85 x

C2 / Nm2)(15,000 N / C)4~(0.17m)* = 4.82 x lo-* C

This value includes the charge on the inner sphere, the charge on the inside surface of the hollow sphere, and the charge on the exterior surface of the hollow sphere due to polarization. Thus,

27.40. Model: The charge distribution at the surface of the earth is assumed to be uniform and to have

spherical symmetry.

Visualize:

Gaussian surface

Earth

--_- ---

Due to the symmetry of the charge distribution, E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the surface. Solve: Gauss's law is me = $ E . d i = Q,,/ E , . The electric field points inward (negative flux), hence Q,,, =

= -(8.85 x lo-" C' / N m2)(100N / C ) 4 ~ ( 6 . 3 7 x lo6 rn)' = 4 . 5 1 x l o 5 C

Gauss's LAW

27.41. Model: The electric field inside the box is

27-1 1

Visualize: Please refer to Figure 27.32.

Solve: The charge on the capacitor plates produces a uniform electric field pointing to the right. When the metal box

- E,,,

is added, the field inside the box is

because there is no charge enclosed within the box. The principle of

superposition tells us that E,, = Ecapacimr + Ebx, where Ebx is the electric field due to the charge on the exterior surface of the box. The net surface charge is zero, because the box as a whole is neutral, but the box is polarized by the capacitor's electric field to where the left end of the box is negative and the right end is positive. Since E,, = 6, we're led to conclude that the field due to the charge on the surface of the box is EbX= - EC,,,, = a uniform field pointing to the left. Note that this is the electric field inside the box due to the surface charge on the box. The field outside the box is much more complex. If we freeze the surface charges and remove the box from the capacitor, we're left with Eaet = EbX= a uniform field pointing to the left. This is shown in the picture below.

27.42. Model: The hollow metal sphere is charged such that the charge distribution is spherically symmetric.

Visualize:

The figure shows spherical Gaussian surfaces at r = 4 cm, at r = 8 cm, and at r = 12 cm. These surfaces match the symmetry of the spherical charge distribution. So E is perpendicular to the Gaussian surface and the field strength has the same value at all points on the Gaussian surface. Solve: The charge on the inside surface is

Qlnslde = (-100 nC/m2)4n (0.06 m)' = -4.524 nC This charge is caused by polarization. That is, the inside surface can be charged only if there is a charge of 4 . 5 2 4 nC at the center which polarizes the metal sphere. Applying Gauss's law to the 4.0cm-radius Gaussian surface, which encloses the 4 . 5 2 4 nC charge,

- -

Q,, = $ E . dA =

Q," EAsphere =EO

* E = - =Q,. AE,

4.524

10-9

4n~,,(0.04 m)'

4.524 x

c(9.0 x lo9 N m2 / c') (0.04 m)'

At r = 4 cm, ?i = (2.54 x 10" N / C. outward).

=2.54X1O4 N / C

- -

Thew is no electric field inside a conductor in electrostatic equilibrium. So at r = 8 cm, E = 0. Applying Gauss's law tc - 3cm-radius sphere, =

Q," $ E . d2 = EArphew= EO

27-12

Chapter 27

The charge on the outside surface is Qo,,,d, = (+lo0 nC/m2)4z (0.10 m)'= 1.257 x lo-* C 3 Q,

= Qourslde +

= 1.257 x lo-* C

* E = - =Q,"

- 0.452 x lo-' C + 0.452 X lo4 C = 1.257 X lo* C

1.257 x lo* C = 7.86 x io3 N (8.85 x lo-'' C2 / Nm2)4z(0.12m)'

&,A

= (7.86 x lo3 N / C, outward).

At r = 12 cm,

27.43. Model: The charged hollow spherical shell is assumed to have a spherically symmetric charge distribution. Visualize:

,, ~

-I

The figure shows two spherical Gaussian surfaces at r < R and r > R. These surfaces match the symmetry of the spherical charge distribution. So, l? is perpendicular to the Gaussian surface and the field strength has the same value at all points on the surface. Solve: (a) Gauss's law applied to the Gaussian surface inside the sphere (r < R) is

(b) Gauss's law applied to the Gaussian surface outside the sphere ( r > R ) is

Assess: A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the center of the shell. Additionally, the electric field for r < R is zero, meaning that the shell of charge exerts no electric force on a charged particle inside the shell.

27.44. Model: The hollow plastic ball has a charge uniformly distributed on its outer surface. This distribution leads to a spherically symmetric electric field. _--_ Visualize: Charge Q on surface I I

-___-

surfaces

The figure shows Gaussian surfaces at r < R and r > R.

Gauss's Law

27-13

Solve: (a) Gauss's law for the Gaussian surface for r < R where Q, = 0 is

me = f i . d A = g = O

Nm2/C * E = O N / C

(b) Gauss's law for the Gaussian surface for r > R is

Assess: A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the center of the sphere. Additionally, the shell of charge exerts no electric force on a charged particle inside the shell.

27.45. Model: The charge distributions of the ball and the metal shell are assumed to have spherical symmetry. _ _ - - - - _-_

Visualize:

Spherical Gaussian surfaces

The spherical symmetry of the charge distribution tells us that the electric field points radially inward or outward. We will therefore choose Gaussian surfaces to match the spherical symmetry of the charge distribution and the field. The figure shows four Gaussian surfaces in the four regions: r 5 a, a < r < b, b I r Ic and r > c. Solve: (a) Gauss's law is me = fE.d;i . Applying it to the region r I a, where the charge is negative so

=e,/â&#x201A;Ź,

E points inward, we get -EAspkx = -E(4m2) = + p ( q r 3 )* E = EO

-Pr 3g0

Here p = - Q / F u 3 is the charge density (Urn3). Thus

Applying Gauss's law to the region a < r < b, '

-EAsphex =

E = -- Q,,=--~ Z E , r7

1 Q, 4 m 0 r'

?i = 6 in the region b Ir Ic because this is a conductor in electrostatic equilibrium. To apply Gauss's to the region r > c, we use Q,, = -Q + 2Q = +Q. Thus,

27-14

Chapter 27

27.46. Model: The three planes of charge are infinite planes. Visualize:

iEp,, Region 1

Region 2

P' ZP

, ~

Region 3

From planar symmetry the electric field can point straight toward or away from the plane. The three planes are labeled as P (top), P', and F"'(bottom). Solve: From Example 27.6, the electric field of an infinite charged plane of charge density q is

In region 1 the three electric fields are

Adding the three contributions, we get I?,,, = 6 N I C. In region 2 the three electric fields are

Thus, E",, = (7 7 / 2 E 0 ) j . In region 3,

Thus,

E",

= -(77/2&,)j.

Gauss's Law

27-15

In region 4,

Thus E,,, = 6 N / C .

27.47.

Model: The charge has planar symmetry, so the electric field must point toward or away from the slab. Furthermore, the field strength must be the same at equal distances on either side of the center of the slab. Visualize:

z >i o

Choose Gaussian surfaces to be cylinders of length 22 centered on the z = 0 plane. The ends of the cylinders have area A . Solve: (a) For the Gaussian cylinder inside the slab, with z < z,, Gauss's law is E.di+j

$E.dA=J E.dA+j bottom

tOP

sides

E.&=% Ell

The field is parallel to the sides, so the third integral is zero. The field emerges from both ends, so the first two integrals are the same. The charge enclosed is the volume of the cylinder multiplied by the charge density, or Q,, = poV = p0(2zA).Thus E.di+j

%L-

I,p

E . & + O = 2 E A * E = LP Z

bottom

The field increases linearly with distance from the center. (b) The analysis is the same for the cylinder that extends outside the slab, with charge Q = p0(2z,,A)is that within a cylinder of length 2% rather than 2z. Thus -EO

=IopE.di+j EO

bottom

E.&+O=2EA=,E=-

> q,except that the enclosed POZO Ell

The field strength outside the slab is constant, and it matches the result of part (a) at the boundary.

27-16

Chapter 27

27.48. Model: The infinitely wide plane of charge with surface charge density q polarizes the infinitely wide conductor. Visualize: Region 1 Zp

Region 3

",YE:

E;.

P" (q)

4';

Region4

EP'

Because E = 6 in the metal there will be an induced charge polarization. The face of the conductor adjacent to the plane of charge is negatively charged. This makes the other face of the conductor positively charged. We thus have three infinite planes of charge. These are P (top conducting face), P'(bottom conducting face), and F"'(p1ane of charge). Solve: Let q,, q2,and q3 be the surface charge densities of the three surfaces with q2 a negative number. The electric field due to a plane of charge with surface charge density 17 is E = 1 7 / 2 .~Because ~ the electric field inside a conductor is zero (region 2),

E, +E,.

1 7 j- +- 7 7 2 j- +- v 3j- = 0- N / C * -17, +q, + 17 = 0 C / m2 + Ep,,= 0 N / C * -2 2Eo

2E0

We have made the substitution q3= 77. Also note that the field inside the conductor is downward from planes P and P' and upward from P".Because q1+ q2= 0 C/m2,because the conductor is neutral, q2= -ql. The above equation becomes - q l - q , + q = O C / m m 2*ql =+17*q2 =-+q We are now in a position to find electric field in regions 1-4. For region 1,

The electric field is

~n region 2,

E,,, = E, + E,. + E,,, = (17/2&,)3.

E,,, = 6 N/C. ~nregion 3,

The electric field is E,,, = (77/2E0)j In region 4,

The electric field is E,,, = -( t7/2&,)3.

Gauss’s Law

21-11

27.49. Model: Assume the metal slabs are large enough to model as infinite planes. Then the electric field has planar symmetry, pointing either toward or away from the planes. . Visualize:

b C

One Gaussian surface is a cylinder with end-areas a extending past the two metal slabs. A second Gaussian surface ends in region 3, the space between the slabs. Solve: (a) Because these are metals, we immediately know that i2= E, = 6. To find the fields outside the slabs, consider the Gaussian surface on the left. The electric field everywhere points up or down, so there is no flux through the sides of this cylinder. Because both slabs are positive, the electric fields in regions 1 and 5 point outward. Gauss’s law is

The Gaussian surface encloses both the upper and lower surfaces of the top slab. The total charge per unit area on both surfaces of this slab is Q,/A = QIA, so the charge enclosed within the cylinder is QalA. (For this calculation we don’t need to know how the charge is distributed between the two surfaces.) Similarly, the enclosed charge on the lower slab is Q2a/A= 2QaIA. Thus E , a + E , a = -Qa + - a E2Qa ,+E, €,A €,A

=-3Q

€,A

Fields and E, are both a superposition of the fields of four sheets of surface charge. Because the field of a plane of charge is independent of distance from the plane, the superposition at points above the top plane must be the same magnitude, but opposite direction, as the superposition at points below the bottom plane. Consequently, E, = Es (same field strengths). This is a rather subtle point in the reasoning and one worth thinking about. Thus E, = E, = 3Q/2@. Now consider the Gaussian surface on the right. The lower slab is more positive than the upper slab, so the electric field in region 3 must point upward, into the lower face of this cylinder. Thus

where the minus sign with E3 is because of the direction. We know E,, and we’ve already determined that Q,, = Qa/A. Thus

E = E --=---.=Q 3 Q Q Q 3 1 €,A 2€,A &,A 2&,A Summarizing, E , = 3Q12qA. E: = 0. E, = Q/9@. E, = 0, and E, = 3Q/7~&. (b)The electric field at the surface of a conductor is E = q/&. We can use the known fields and 17 = .@ to find the four surface charge densities. At surface a, E, points away from the surface. Thus 77, = +€,E, = + E , At surface b,

E, points

3Q 3Q - +-2A 2€,A

toward the surface. Thus

Surface c is opposite to surface b, because the field points away from the surface, so q, = +Q/2A. Finally, at surface d , the field points away from the surface and has the same strength as E,, hence qd= +3Q/2A. Assess: Notice that 17, + vb= Q/A and 77, + qd= 2Q/A. This is the expected “net” surface charge density for slabs with total charge Q and 2Q.

27-18

Chapter 27

27.50. Model: A long, charged wire can be modeled as an infinitely long line of charge. +A visualize:

Gaussian surfaces

c End view

The figure shows an infinitely long line of charge that is surrounded by a hollow metal cylinder of radius R. The symmetry of the situation indicates that the only possible shape of the electric field is to point straight in or out from the wire. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius r and length L, centered on the wire. Solve: (a) For the region r < R, Gauss’s law is

aL a.L. ~ O N m 2 / C + O N m 2 / C + E . A s =-e. ,, E(2m)L=-

‘ ’ - [2iEo :,

*E=--

2m0 r

*E=

h outward) = --

~zE,, r

(b) Applying Gauss’s law to the Gaussian surface at r > R,

27.51.

Model: A long, charged cylinder is assumed to be infinite &d to have linear symmetry.

Visualize:

-L-

- _ -

Cylindrical Gaussian surfaces

The cylindrical symmetry of the situation indicates that the only possible shape of the electric field is to point straight in or out from the cylinder. The shape of the field suggests that we choose our Gaussian surface to be a cylinder of radius rand length L, which is concentric with the charged cylinder.

Gauss’s Law

27-19

Solve: (a) For r 1 R, Gauss’s law is

Q,, aL 0 N m’ / C + O N m2 / C + EA = -3 E = = -* , E = ( 2 m - L ) ~ ~2 m 0 r EO

2m0 r

(b) For r I R, Gauss’s law is m = - *QhE = - = -

Qill

A&,

P(m-’L) (2m-L)~~

An expression for the volume charge density p in terms of the linear charge density can be calculated by considering the charge on a cylinder of length d and radius R:

I aR, 1 a ErZR= -2nEoR? = --r 2m0 R

in the shell has spherical symmetry. 27.52. Model: The charge distribution _---_

Visualize:

i \ 5

! 1

I I

I I I I I I I

--_--_-Spherical Gaussian surfaces

The spherical surfaces of radii r L R,,, r 5 R,, and R,, I r I R,,, concentric with the spherical shell, are Gaussian surfaces. Solve: (a) Gauss’s law for the Gaussian surface r 2 R, is

Q,” jE ( 4 m - ) = -jE = L f E- . dA- = 9

1 Q E- = -~ Z E r’ ,

4 m 0 r’

The vector form comes from the fact that the field is directed radially outward. (b) For r 5 R,,, Gauss’s law is f E . d i = - -Q-, , - O C Eo

” I

... ..

E- = O-

;

Chapter 27

27-20

(c)For R, I r I Ram, Gauss’s law is

Qf4ns&&,

r’Rau1

27.53. Model: Assume that the negative charge uniformly distributed in the atom has spherical symmetry, Visualize:

Unifody distributed charge

Spherical Gaussian surface

Point charge

The nucleus is a positive point charge +Ze at the center of a sphere of radius R. The spherical symmetry of the charge distribution tells us that the electric field must be radial. We choose a spherical Gaussian surface to match the spherical symmetry of the charge distribution and the field. The Gaussian surface is at r < R, which means that we will calculate the amount of charge contained in this surface. Solve: (a) Gauss’s law is f E . d;i = Q,,/ E ~ The . amount of charge inside is

(b)At the surface of the atom, r = R. Thus,

Gauss's Law

27-21

This is an expected result, which can be quickly obtained from Gauss's law. Applying Gauss's law to a Gaussian surface just outside r = R . Because the atom is electrically neutral, Q,"= 0. Thus

, j ~ . d i i = L o = E, = O N / C EO

E,, = 92(1.60 x l O - I 9 C)(9.0 x lo9 C2 / Nm')

- (0*05nm) = 4.64 x 1013N / c (0.05 m)* (0.10 MI)^

27.54. Model: The long thin wire is assumed to be an infinite line of charge. Visualize: Please refer to Figure CP27.54. The cube of edge length L is centered on the line charge with a linear charge density A. Although the line charge has cylindrical symmetry, we will take the cube as our Gaussian surface. Solve: (a) The electric flux through an area d i in the yz plane is d@ = E . d i . The electric field E due to an infinite line of charge at a distance s from the line charge is

Also

d;i = L d y l . Thus, 1

d@=- A

Ldy( F .l) =

-d

cos0

Y +(L/2)

nL dy

nL' dy

LJ2

J+(L/2)2 = 4 ~ ~ 0 [ Y 2(L/2)']

(b)The expression for the flux d@ can now be integrated to obtain the total flux through this face as follows:

( c ) Because there are four faces through which the flux flows, the net flux through the cube is

27.55. Model: The field has cylindrical symmetry. Visualize: R

Gaussian

'surface -x

-R

Solve: (a) The volume charge density p ( r ) = r p o / R is linearly proportional to r. The graph is shown above. OD) Consider the cylindrical shell of length L, radius r and thickness dr shown in the diagram. The charge within a small volume dV is

P' 7"p L d q = pdV = L ( 2 n r ) d r L = -r2dr R R

27-22

Chapter 27

Integrating this expression to obtain the total charge in the cylinder:

< R shown in the figure. Gauss's law is

= $2'd i =

Q, / E .~The charge on the inside of the Gaussian surface is

The last step is justified because the electric field has radial dependence. (a)The expression for the electric field strength simplifies at r = R to 2 E=--=-A R' 2 m 0 R3 2n&,R

This is the same result as obtained in example 27.5 for a long, charged wire.

27.56. Visualize:

@@ ,'.--' ,

Gaussian Spherical surface

Solve: (a) Consider the spherical shell of radius r and thickness dr shown in the figure. The charge d q within a small volume dV is C dq = pdV = - ( 4 m 2 ) d r = 4nCdr r2 Integrating this expression to obtain the total charge in the sphere: R

Q=jdq=j4Kdr=4KR

* C=- Q

4zR (b) Consider the spherical Gaussian surface at r < R shown in the figure. Gauss's law applied to this surface is 0

Using the results from part (a), = b ( & ) n d r = ~ rQ Q,, = j d q = j p dV = j-;4m2dr ' C 0 r-

E=--1

4 7 R'~ ~

This is the same result as obtained in Example 27.3. The result was expected because a spherical charge behaves, for r t R , as if the entire charge were at the center.

Gauss’s Law

21-23

Model: The charge density within the sphere is nonuniform. We will assume that the distribution has spherical symmetry. visualize:

27.57.

@ \

-_-,

Spherical Gaussian surface

Solve: (a) Consider the thin shell of width dr at a distance r from the center shown in the figure. The volume of this thin shell is dV = ( 4 m ’ ) d r . The charge contained in this volume is

(b) Gauss’s law for the Gaussian surface shown in the figure at r < R is

The charge inside the Gaussian surface is

Gauss’s law becomes

The direction of the field is radially outward. (c) At r = R, the above expression for the electric field reduces to

E = &(4 = -1 Q 4m0R3 ~ ? L ER2 ~ This is an expected result, because all charge Q is inside R and the field looks like that of a point charge. 27.58.

Visualize:

, -_.-

Spherical Gaussian surface

Solve: (a) It is clear that E(r) oc r4 up to r = R. That is, the maximum value of E(r) occurs at r = R. At this point,

E=-- 1 Q 4m0 R’ because all the charge is inside R and the charge is spherically distributed about a point at the center. Thus,

(b) Applying Gauss’s law to the surface at r < R shown in the figure,

27-24

Chapter 27

To find Qh, we consider the thin spherical shell of width dr and charge dq at a distance r from the center shown in the figure. Thus,

dq = pdV = p ( 4 m 2 ) d r

* J d q = Q,

=jp(r)(4m2)dr

Using this form of Q, in the equation obtained from Gauss’s law,

Q r6 * j p ( r ) r 2 d r= -47c R6

Q,,, = Q?r6 = j p ( r ) 4 m 2 d r R Taking the position derivative on both sides

d J p ( r ) r 2 d r= Q - (r6) dr

4nR6 dr

Q (6.’) p(r)r2 = 4ZR6

3Qr3 p ( r )= -

2nR6

3Qr3 dq = p(r)dV = - 4 m 2 d r 2ZR6

6Q r5 * dq = 6 Q -R6 d r * Q = Idq = B r r ’ d r = -.R6 R6 0

R6 6