engineeringhero.files.wordpress.com27-4 chapter 27 solve: the electric field is parallel to four...
TRANSCRIPT
GAUSS'S LAW
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: t
+++++++++++++++++
+++++++++++++++++ 1 ! 1 1 ! 1 1 1 1 1 1 1 1 1 1
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. positive charge and inro a closed surface surrounding a net negative charge. Visualize: Solve: line perpendiculur to the plane of the surface. The electric flux our of the closed cube surface is
Model: The electric f lux "flows" our of a closed surface around a region of space containing a net
Please refer to Figure Ex27.4. Let A be the area in m2 of each of the six faces of the cube. The electric flux is defined as = €.A = EAcose, where 8 is the angle between the electric field and a
(Pa", = (20 N / C + 20 N / C + 10 N /C)AcosO" = (50A) N m' / C Similarly. the electric t lux inro the closed cube surface is
= (15 N / C + 15 N / C + 15 N / C)Acos180° = - ( U A ) N m' / C The net electric flux is (50A) 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. 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: 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+lON / C + 5 N / C)AcosO" = (35A) N m2 / C
Model: The electric flux "flows" out of a closed surface around a region of space containing a net
The electric flux is defined as Qe = E . 2 = EAcose, where 8 is the angle between the electric field and a
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. 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 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
Similarly, the electric flux into the closed cube surface is
Because the cube contains negative charge, Qom + Qm must be negative. This means Qom + Q," + Therefore,
Model: The electric flux "flows" out of a closed surface around a region of space containing a net
Qout = (20 N / C + 10 N / C + 10N / C)AcosO" = (40A) N m2 / C
Q,,, = (20 N / C + 15 N / C)Acos180° = -(35A) N m2 / C
< 0 N m2/C.
(40A) N m2 / C+(-35A) N m2 / C+Qbo,, < 0 N m2 / C
* Qbow < (-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. positive charge and into a closed surface surrounding a net negative charge. Visualize: Solve: line perpendicular to the plane of the surface. The electric flux out of the closed cube surface is
Similarly, the electric flux into the closed cube surface is
Because the cube contains no charge, the total flux is zero. This means
Model: The electric flux "flows" out of a closed surface around a region of space containing a net
Please refer to Figure Ex27.7. Let A be the area of each of the six faces of the cube. The electric flux is defined as Qe = E.2 = EAcos8, where 8 is the angle between the electric field and a
QoUl = ( 1 5 N / C + 1 5 N / C + 10N/C)Acos0°=(40A)Nm2/C
a,,, = (20 N / C + 15 N / C)Acos180° = -(35A) N m2 / C
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. Visualize: Because the normal
Solve: The electric flux is
Model: The electric field is uniform over the entire surface. 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 e = 600.
Q e = E . i =EAcos8=(200N/C)(1.0~10-~ m2)cos60"=1.0Nm2 / C
27.9. Visualize: the normal Solve: The electric flux is
Model: The electri'c field is uniform over the entire surface. Please refer to Figure Ex27.9. The electric field vectors make an angle of 30" below the surface. Because
and E is 8= 120". to the planar surface is at an angle of 90" relative to the surface, the angle between
oe = E - i = m c o s e = (200 N cl(1.0 x IO-? mz)cos12~o = -1.0 N m' / c
Gauss's Law 27-3
27.10. Model: 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
The electric field is uniform over the entire surface.
@e = E . ~ = E A c o s O ~ 1 5 . 0 N m 2 / C = E(l.OxlO-' m 2 ) c o s 3 0 " ~ E = 1 7 3 0 N / C
27.11. Model: Solve:
The electric field is uniform over the rectangle in the xy plane. (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
De = E.2 = (50; + lOOi) .(6.0 x l0"i) N m2 / C = 6.0 x lo-' N m' / C
me = E . 2 = ( 5 0 1 + 100~)~(6.0x10'k I N m ' / C = O N m ' / C
The electric flux through the rectangle is
(b) The electric flux is
Assess: In (b), E is in the plane of the rectangle. That is why the flux is zero.
27.12. Solve:
Model: (a) The area vector is perpendicular to the xz plane. Thus
The electric field over the rectangle in the xz plane is uniform.
2 = (2.0 cm x 3.0 cm)j = (6.0 x lo4 m 2 ) j
The electric flux through the rectangle is
@ e =,??-2=(501 + 100i)N/C~(6.Ox1O4 m 2 ) j
=(3O0x1O4 N r n 2 / C ) ( ~ . ~ ) + ( 6 0 0 x 1 O 4 N m ' / C ) ( ~ . ~ ) = O N m * / C
(b) The flux is
@e = E . 2 = (50; + l00j) N / C.(6.0 x lo4 m') j
=(3O0x1O4 N m 2 / C ) ( ~ . ~ ) + ( 6 0 0 x 1 0 4 N m 2 / C ) ? . j
= 0 N m2 /C(0.06 N m2 / C) = 0.06 N m2 / C
27.13. Solve:
Model: The area vector of the circle is
The electric field over the circle in the xy plane is uniform.
2 = nr2i = ~ ( 0 . 0 2 m ) 2 i = (0.001257 m 2 ) i
me = E . i =( lOOO~+lOOO~+lOOO~)N/C-(0 .001257 m')L
Thus, the flux through the area of the circle is
A - 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: Visualize: - - -
The electric field over the two faces of the box perpendicular to E is uniform.
- - - 4 4
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: Visualize: electric field strength.
Solve: end, where E' points outward, Altogether, the net flux is @e = 0 N m'/C2.
(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.
The electric field through the two cylinders is uniform. Please refer to Figure Ex 27.15 . Let A = nR2 be the area of the end of the cylinder and let E be the
(a) There's no flux through the side walls of the cylinder because E' is parallel to the wall. On the right
= EA = nR2E. The field points inward on the left, so @,er, = -EA = -rR2E.
= EA = nR2E. Thus the --*
27.16. Solve: For any closed surface enclosing a total charge Qm, the net electric flux through the surface is
me =- 'In * Q, = gome = (8.85 x C2 / Nm2)(-1000 N m2 / C) = -8.85 nC EO
27.17. Solve: For any closed surface that encloses a total charge Qh, the net electric flux through the surface is
Q. (55.3 X 106)(-1.60 X lo-'' c) @ =a= = - 1 . 0 N m 2 / C
e €0 8.85 x lo-" C2 / N m2
27.18. Solve: surface is me = Q,,, /E, . The flux through the top surface of the cube is one-sixth of the total:
For any closed surface that encloses a total charge Q,,, the net electric flux through the closed
= 188 N m* / C - Qin - 10 x io-' c 6 ~ , 6(8.85 x lo-'* C2 / N m2) @ e slllface
27.19. Solve: surface is me = ',,,/E, . The total flux through the cube is
For any closed surface that encloses a total charge Qin, the net electric flux through the closed
@e
27.20.
= 6(100 N
Visualize:
m2
I
= (600
0 4
N m2 / C)(8.85 x lo-'' C2 / N m2) = 5.31 x ~ o - ~ c = 5.31 nC
, : ,o I , , ' _ - oq
(Cf (4 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:
- 1 ~ 1 0 - ~ c Qe = = -113 N m2 / C
8.85 x lo-” C2 / Nm2 This is inward flux.
27.23. Visualize: electric flux through the closed surface is @ e = +lo0 nC and the -100 nC charges are outside the cylinder. Thus,
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
= 113 N m2 / C. 1 . 0 ~ 1 0 - ~ c
Qe = 8.85 x lo-’’ Cz / Nm2
This is outward flux.
27.24. Model: Solve:
A copper penny is a conductor. The excess charge on a conductor resides on the surface. The electric field at the surface of a charged conductor is
q = E ~ E ~ , ~ ~ ~ = (8.85 x lo-’’ C 2 / Nm2)(2000 N / C) = 17.7 x C / m2
27.25. Solve:
Model: The electric field at the surface of a charged conductor is
The excess charge on a conductor resides on the outer surface.
perpendicular to surface
j q = E,E,,,, = (8.85 x lo-” C’ / Nm2)(3 x 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. Visualize: Solve:
Model:
Point 1 is at the surface of a charged conductor, hence
The excess charge on a conductor resides on the outer surface. Please refer to Figure P27.26.
(5.0 x 101’)(1.60 x C/m‘)
8.85 x IO-’’ C’ / N m’ = 904 N / C perpendicular to surface
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. Visualize: Solve: electric field and the vector ;? that points outward from the surface. Thus
Model:
(a) The electric flux through a surface area 2 is Qe = i?.i = EAcos8, where 8 is the angle between the
The electric field over the four faces of the cube is uniform. Please refer to Figure €27.27.
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 Qe = Q, + Q2 + Q3 + 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: Visualize: 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
The electric field over the five surfaces is uniform. Please refer to Figure P27.28.
0, = E,A, cos8, = (500 N / C)(1 m x 2 m)cos(l80") = -1000 N m2 / C
Q2 = E 2 A , c o s 9 0 0 = Q , = Q 5 = O N m 2 / C
Q, = E4A4 cos8, = (500 N / C)((l m/sin30") x 2 m)cos60" = +lo00 N m2 / C
Assess: found.
Because the flux into these five faces is equal to the flux out of the five faces, the net flux is zero, as we
27.29. Visualize:
Model: Because the tetrahedron contains no charge, the net flux through the tetrahedron is zero. * E
- + - - .15 h = a cos 60" = - a - - - 1 4/+ - - I ' I d , I 60" a Abase = 1 . 1 5 ah = - a' 2
4
+
Abase
Solve: (a) The area of the base of the tetrahedron is +aZ, 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
4 4
(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: CDe = Q, / E , . We can write three equations from the three closed surfaces in the figure:
For any closed surface that encloses a total charge Q,", the net electric flux through the surface is
Subtracting third equation from the first,
41 - 92 = +9
2q, = +4q 3 q1 = 2q
Adding second equation to this equation,
That is, q1 = +2q, q2 = +q, and q3 = -3q.
27.31. electric flux through the closed surface is
Solve: The sphere encloses both q, and q2. For any closed surface that encloses a total charge Q,", the net
Q,,, - (-4Q + 2Q) - -- -2Q @ =-- e Eo EO EO
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=101 N m 2 / C
Q,. = qCPe = (8.85 x lo-'' C2/N m2)(100.5 N m2/C) = 8.90 x lo-'' C
(c) Because CDe = e,,/&,, ,
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, = $2 . d i = EAsphere = (2000 N / C)( 0.1 257 m ' ) = 25 1 N m' / C
(c) Because @e = Q,. / E ~ ,
Q,, = qCDe = (8.85 x lo-'' C'/N m2)(251 N mZ/C) = 2.22 x C = 2.22 nC
27.34. Solve: copper plate, so the surface charge density is
Model: The copper plate is a conductor. The excess charge resides on the surface of the plate. (a) One-half of the electrons are located on the top surface and one-half on the bottom surface of the
(0.5x10'0)(1.60x10~'9 C)
0.1 m x 0.1 m 7 7 = = 8 0 ~ 1 0 - ~ C / m '
Thus, the electric field at the surface of the plate is 17 80x10" C / m '
E, E = - = = 9 0 4 0 N / C
8.85 x lo-'' C' / N m2 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.
27-8 Chapter 27
2735. Model: The excess charge on a conductor resides on the outer surface. Visualize: Before After
, + 100 nC
\
Gaussian surface 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 Eo
@ =-
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
3(80 x c) p = - - - Q - Q = = 2.39 x 10" C / m3
V Sm-' 4 ~ ( 0 , 2 0 m ) ~ (b) Once again, because the ball is uniformly charged,
Q = pV = p( $ r 3 ) = (2.387 x 10" C / m3) - = (1.00 x lo-' C / m')r7 ("3" m e n r = 5 c m , Q5 = ( l . O 0 ~ 1 0 - ~ C/m')(0.05m)3=1.25nC.Whenr=10crnand20cm, Q l o = l O . O n C ~ d Qro= 80.0 nC .
GUUSS’S Law 27-9
(c) Gauss’s law is @ e = f E . d i = EA = Q, / E ~ . For the 5.0-cm-radius Gaussian spherical surface,
= 4500 N / C 1 . 2 5 ~ 1 0 - ~ C = EsAs j E5 = - Q5 - - EO EoAs (8.85 x IO-” C2 / Nm’)[4~(0.05 m)’]
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. Assess: We note that = 2EI0 = 4Es. This result is consistent with the result of Example 27.4 according to which
27.38. Model: within the hollow metal sphere has spherical symmetry. Visualize:
The excess charge on a conductor resides on the outer surface. The field inside, outside, and
E t
- Spherical E 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,” EO
- - @ e = f E . dA = EAspherc = E( 4 d ) = -
Here Q,, = 0 C. I t 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 + (-e) + (+e) = +3Q
27-10 Chapter 27
Gauss's law applied to the Gaussian surface at r 1 b yields:
Qi,, - +3Q 1 3 Q - 1
E, Eo
. . . - me = 4 E . dA = EAsphere = E(4m2) = - - - j E = -( T ) a E = -( y);
4 m 0 4ZEO (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:
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 = fz . dA = e,,,/&, . Applying it to a Gaussian surface of radius 8 cm,
The hollow sphere still has the same charge +2Q as given in the problem, although the sphere is polarized.
Q,,, = -E,EA,~~~, = -(8.85 x C2 / N m2)(15,000 N / C)[4~(0.08 m)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, Qin is 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 ~ $ E . dA = E,EA,,, = (8.85 x C2 / Nm2)(15,000 N / C)4~(0.17 m)* = 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: spherical symmetry.
The charge distribution at the surface of the earth is assumed to be uniform and to have
Visualize: A [u I 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)(100 N / C)4~(6.37 x lo6 rn)' = 4 . 5 1 x lo5 C
Gauss's LAW 27-1 1
27.41. Model: The electric field inside the box is 6. Visualize: Solve: The charge on the capacitor plates produces a uniform electric field pointing to the right. When the metal box is added, the field inside the box is E,,, = 6 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
Please refer to Figure 27.32.
- -
Eaet = EbX = a uniform field pointing to the left. This is shown in the picture below.
27.42. Model: Visualize:
The hollow metal sphere is charged such that the charge distribution is spherically symmetric.
i
E 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 .524 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," EO
- - Q,, = $ E . dA = EAsphere = -
4.524 10-9 c 4.524 x c(9.0 x lo9 N m2 / c') - - =2.54X1O4 N / C * E = - = Q,.
AE, 4n~,,(0.04 m)' (0.04 m)'
At r = 4 cm, i? = (2.54 x 10" N / C. outward).
law tc - 3cm-radius sphere,
- - Thew is no electric field inside a conductor in electrostatic equilibrium. So at r = 8 cm, E = 0. Applying Gauss's
= $ E . d2 = EArphew = - Q," 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 - 0.452 x lo-' C + 0.452 X lo4 C = 1.257 X lo* C
= 7.86 x io3 N c Q," 1.257 x lo* C
&,A * E = - =
(8.85 x lo-'' C2 / Nm2)4z(0.12 m)'
At r = 12 cm, = (7.86 x lo3 N / C, outward).
27.43. Visualize:
Model: The charged hollow spherical shell is assumed to have a spherically symmetric charge distribution.
,, ~ - -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. leads to a spherically symmetric electric field. Visualize:
Model: The hollow plastic ball has a charge uniformly distributed on its outer surface. This distribution
_ - - _
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 EO
(b) Gauss's law for the Gaussian surface for r > R is
Assess: 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:
A uniform spherical shell of charge has the same electric field at r > R as a point charge placed at the
_ _ - - - - _ _ -
I
,
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 I c and r > c. Solve: (a) Gauss's law is me = fE.d;i =e,/€, . Applying it to the region r I a, where the charge is negative so
E points inward, we get
-EAspkx = -E(4m2) = + p ( q r 3 ) * E = - - Pr EO 3g0
Here p = - Q / F u 3 is the charge density (Urn3). Thus
Applying Gauss's law to the region a < r < b,
' -EAsphex = 9 j E = -- ' Q,,=--- 1 Q, EO ~ Z E , r7 4 m 0 r'
i? = 6 in the region b I r I c 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
E
27.46. Model: The three planes of charge are infinite planes. Visualize:
Region 1
EP iEp,, P
Region 2
P'
Z P , ~ 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: Furthermore, the field strength must be the same at equal distances on either side of the center of the slab. Visualize:
The charge has planar symmetry, so the electric field must point toward or away from the slab.
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 . d A = J E . d A + j E . d i + j E.&=% tOP bottom sides
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
%L- E . d i + j E . & + O = 2 E A * E = L P Z
-- - = I o p E . d i + j E .&+O=2EA=,E=- P O Z O
EO I,p bottom EO EO
The field increases linearly with distance from the center. (b) The analysis is the same for the cylinder that extends outside the slab, with 2 > q, except that the enclosed charge Q = p0(2z,,A) is that within a cylinder of length 2% rather than 2z. Thus
Ell bottom EO EO
The field strength outside the slab is constant, and it matches the result of part (a) at the boundary. (4 E
27-16 Chapter 27
27.48. Model: conductor. Visualize:
The infinitely wide plane of charge with surface charge density q polarizes the infinitely wide
Region 1 Zp
Region 3
P" (q)
",YE:
4'; Region4
EP' E;.
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),
1 7 - 7 7 2 - v 3 - - 2Eo 2E0 2 E 0
- - - - E, +E,. + Ep,, = 0 N / C * -2 j +- j +- j = 0 N / C * -17, +q, + 17 = 0 C / m2
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 E,,, = E, + E,. + E,,, = (17/2&,)3.
~n region 2, E,,, = 6 N/C. ~n region 3,
The electric field is E,,, = ( 7 7 / 2 E 0 ) j In region 4,
The electric field is E,,, = -( t7/2&,)3.
Gauss’s Law 21-11
27.49. Model: planar symmetry, pointing either toward or away from the planes. Visualize:
Assume the metal slabs are large enough to model as infinite planes. Then the electric field has
- ..
a
b
C
d
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 = - + - a E , + E , Qa 2Qa =- 3Q €,A €,A €,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 € ,A 2€,A &,A 2&,A 3 1
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
3Q - +-- 3 Q 2€,A 2 A
77, = +€,E, = + E , ~ -
At surface b, E, points 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 visualize: +A
be modeled as an infinitely long line of charge.
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 ( 2 m ) L = - EO EO
h i 2 m 0 r ~ z E , , r
* E = - - ‘ ’ * E = - [2iEo -- :, outward) = --
(b) Applying Gauss’s law to the Gaussian surface at r > R,
27.51. Visualize: -L-
Model: A long, charged cylinder is assumed to be infinite &d to have linear symmetry.
- _ -
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
* , E = -- 2 m 0 r
Q,, aL 0 N m’ / C+O N m2 / C + EA = - 3 E = - = -
( 2 m - L ) ~ ~ 2 m 0 r EO
(b) For r I R, Gauss’s law is
Qh Qill P(m-’L) m = - * E = - = - EO A&, ( 2 m - 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:
(c) At r = R,
I aR, 1 a A ErZR = -- 2nEo R? = --r
2 m 0 R
27.52. Visualize:
Model: The charge distribution in the shell has spherical symmetry. _ - - - _
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,” 9 Q - 1 Q ; f E . dA = - j E ( 4 m - ) = - j E = L e j E = -- EO EO 4m0 r’ ~ Z E , r’
The vector form comes from the fact that the field is directed radially outward. (b) For r 5 R,,, Gauss’s law is
fE .d i= - - - Q , , - O C j E = O - - Eo Eo
._ .. I” ... ..
27-20 Chapter 27
Qf4ns&&, -
(c) For R, I r I Ram, Gauss’s law is
I , r’Rau1
27.53. Model: Visualize: Uni fody
distributed charge
Assume that the negative charge uniformly distributed in the atom has spherical symmetry, 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
(c) The electric field strength at r = 3 R = 0.05 nm is
(0*05 nm) = 4.64 x 1013 N / c 1 - (0.05 m)* (0.10 MI)^
E,, = 92(1.60 x lO-I9 C)(9.0 x lo9 C2 / Nm')
27.54. Model: 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
The long thin wire is assumed to be an infinite line of charge.
Also d;i = L d y l . Thus,
1 dY 1 d@=- A Ldy( F . l) = d- cos0
Y + (L /2)
- nL dy LJ2 nL' dy - 2 K & o J / 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:
P
-x - R 0 R 2R
R
Gaussian ' surface
Solve: 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
(a) The volume charge density p(r ) = rpo/R is linearly proportional to r. The graph is shown above.
P' 7"p L dq = 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:
(c) Consider the cylindrical Gaussian surface of length 1 at r < R shown in the figure. Gauss's law is @ e = $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
A R' 2 2 m 0 R3 2n&,R
E=--=-
This is the same result as obtained in example 27.5 for a long, charged wire.
27.56. Visualize:
@ @ '.--' I Gaussian Spherical ! , , surface
Solve: small volume dV is
(a) Consider the spherical shell of radius r and thickness dr shown in the figure. The charge d q within a
C r2
dq = pdV = - ( 4 m 2 ) d r = 4nCdr
Integrating this expression to obtain the total charge in the sphere: R
Q = j d q = j 4 K d r = 4 K R * C=- Q 0 4zR
(b) Consider the spherical Gaussian surface at r < R shown in the figure. Gauss's law applied to this surface is
Using the results from part (a),
Q,, = j d q = j p dV = j - ;4m2dr ' C = b ( & ) n d r = ~ r Q 0 r-
(c) At r = R, the equation for the electric field obtained in part (b) simplifies to E=--- - 1 Q ;
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
27.57. Model: The charge density spherical symmetry. visualize:
within the sphere is nonuniform.
@ \ - _ - ,
Spherical Gaussian surface
We will assume that the distribution has
Solve: this thin shell is dV = (4m’ )dr . The charge contained in this volume is
(a) Consider the thin shell of width dr at a distance r from the center shown in the figure. The volume of
(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 4m0 R3 ~ ? L E ~ R2
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, = j p ( r ) ( 4 m 2 ) d r
Using this form of Q, in the equation obtained from Gauss’s law,
r6 Q r6 Q,,, = Q? = j p ( r ) 4 m 2 d r * jp (r )r2dr = -- R 47c R6
Taking the position derivative on both sides
3Qr3 (6.’) =j p ( r ) = - Q (r6) p(r)r2 = - d - Jp(r ) r2dr = --
dr 4nR6 dr 4ZR6 2nR6 Q
(c) Consider a spherical shell of width dr and charge dq at a distance r from the center. Then
Q dq = p(r)dV = - 4 m 2 d r * dq = 6Q-dr * Q = Idq = B r r ’ d r = -.- = 3Qr3 r5 6Q R6 2ZR6 R6 R6 0 R6 6