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Physics II
Homework VI
CJ
Chapter 21; 84, 90
Chapter 22; 4, 8, 12, 23, 30, 44
21.84. IDENTIFY: The electric field exerts equal and opposite forces on the two balls, causing
them to swing away from each other. When the balls hang stationary, they are in equilibrium
so the forces on them (electrical, gravitational, and tension in the strings) must balance.
SET UP:
(a) The force on the left ball is in the direction of the electric field, so it must be positive,
while the force on the right ball is opposite to the electric field, so it must be negative.
(b) Balancing horizontal and vertical forces gives qE = T sin /2 and mg = T cos /2.
EXECUTE: Solving for the angle  gives:  = 2 arctan(qE/mg).
(c) As E  ∞,   2 arctan(∞) = 2 (π/2) = π = 180°
EVALUATE: If the field were large enough, the gravitational force would not be important,
so the strings would be horizontal.
21.90. IDENTIFY: Use Eq. (21.7) to calculate the electric field due to a small slice of the line of
charge and integrate as in Example 21.11. Use Eq. (21.3) to calculate F .
SET UP: The electric field due to an infinitesimal segment of the line of charge is sketched
in Figure 21.90.
sin  
cos 
y
x2  y 2
x
x  y2
2
Figure 21.90
Slice the charge distribution up into small pieces of length dy. The charge dQ in each slice is
dQ  Q(dy/a). The electric field this produces at a distance x along the x-axis is dE. Calculate
the components of dE and then integrate over the charge distribution to find the components
of the total field.
dE 
EXECUTE:
1  dQ 
Q  dy 

 2


2 
4 P0  x  y  4 P0 a  x 2  y 2 
Qx 
dy


2
4 P0a   x  y 2 3 / 2 


Q 
ydy

dEy  dE sin   
4 P0a   x 2  y 2 3 / 2 


Qx a
dy
Qx  1
Ex   dEx  

2
2 3/ 2

0
4 P0 a ( x  y )
4 P0 a  x 2
dEx  dE cos 
E y   dEy  
a
Q


2
2 
x  y  0 4 P0 x
y
1
x  a2
2
a
a
Q
ydy
Q 
1
Q 1
1




2


   4 Pa  x 
0
4 P0 a ( x  y 2 )3/ 2
4 P0 a 
x2  y 2 0
x2  a2 
0 
(b) F  q0 E
Fx  qEx 
qQ
1
qQ  1
1

; Fy  qE y 



2
2
2
4 P0 x x  a
4 P0a  x
x  a2 
(c) For x >> a,
Fx  
1  a2 
 1  2 
x 
x2  a2 x 
1
1/ 2
1
a2  1 a2
 1  2    3
x  2x  x 2x
qQ
qQ  1 1 a 2  qQa
, Fy 
  

2
4 P0 x
4 P0 a  x x 2 x3  8 P0 x3
EVALUATE: For
x  a, Fy  Fx
and
F  Fx 
qQ
4 P0 x 2
and
F
is in the –x-direction. For x >> a
the charge distribution Q acts like a point charge.
22.4. IDENTIFY: Use Eq.(22.3) to calculate the flux for each surface. Use Eq.(22.8) to calculate
the total enclosed charge.
SET UP:
E = (5.00 N/C  m) x iˆ + (3.00 N/C  m) z kˆ .
The area of each face is
L2 ,
where L  0.300 m .
nˆ s1 =  ˆj  1  E  nˆ S1 A  0 .
EXECUTE:
nˆ S2 =  kˆ   2  E  nˆ S2 A  (3.00 N C  m)(0.300 m)2 z  (0.27 (N C)  m) z .
 2  (0.27 (N/C)m)(0.300 m)  0.081 (N/C)  m 2 .
nˆ =  ˆj    E  nˆ A  0 .
S3
3
S3
nˆ S4 = kˆ   4  E  nˆ S4 A  (0.27 (N/C)  m) z  0 (since z  0).
nˆ S5 = iˆ  5  E  nˆ S5 A  (5.00 N/C  m)(0.300 m)2 x  (0.45 (N/C)  m) x.
 5  (0.45 (N/C)  m)(0.300 m)  (0.135 (N/C)  m 2 ).
nˆ = iˆ    E  nˆ A  (0.45 (N/C)  m) x  0 (since x  0).
S6
6
S6
(b)Total flux:    2  5  (0.081  0.135)(N/C)  m 2  0.054 N  m 2/C. Therefore,
q  P0  4.78  1013 C.
EVALUATE: Flux is positive when
directed into the volume.
E
is directed out of the volume and negative when it is
22.8. IDENTIFY: Apply Gauss’s law to each surface.
SET UP: Qencl is the algebraic sum of the charges enclosed by each surface. Flux out of the
volume is positive and flux into the enclosed volume is negative.
EXECUTE:
(a) S  q1/P0  (4.00 109 C)/P0  452 N  m2/C.
1
(b) S  q2/P0  (7.80 109 C)/P0  881 N  m2/C.
2
(c) S  (q1  q2 )/P0  ((4.00  7.80) 109 C)/P0  429 N  m2/C.
3
(d) S  (q1  q3 )/P0  ((4.00  2.40) 109 C)/P0  723 N  m2/C.
4
(e) S  (q1  q2  q3 )/P0  ((4.00  7.80  2.40) 109 C)/P0  158 N  m2/C.
5
EVALUATE:
(f ) All that matters for Gauss’s law is the total amount of charge enclosed by the surface,
not its distribution within the surface.
22.12. IDENTIFY: Apply Gauss’s law.
SET UP: Use a small Gaussian surface located in the region of question.
EXECUTE:
(a) If ρ  0 and uniform, then
q inside
any closed surface is greater than zero. This implies
so úE  dA  0 and so the electric field cannot be uniform. That is, since an arbitrary
surface of our choice encloses a non-zero amount of charge, E must depend on position.
0,
(b) However, inside a small bubble of zero charge density within the material with density ρ
, the field can be uniform. All that is important is that there be zero flux through the surface
of the bubble (since it encloses no charge). (See Problem 22.61.)
EVALUATE: In a region of uniform field, the flux through any closed surface is zero.
22.23. IDENTIFY: The electric field inside the conductor is zero, and all of its initial charge lies
on its outer surface. The introduction of charge into the cavity induces charge onto the
surface of the cavity, which induces an equal but opposite charge on the outer surface of the
conductor. The net charge on the outer surface of the conductor is the sum of the positive
charge initially there and the additional negative charge due to the introduction of the
negative charge into the cavity.
(a) SET UP: First find the initial positive charge on the outer surface of the conductor
using qi   A, where A is the area of its outer surface. Then find the net charge on the surface
after the negative charge has been introduced into the cavity. Finally use the definition of
surface charge density.
EXECUTE: The original positive charge on the outer surface is
qi   A   (4 r 2 )  (6.37  106 C/m 2 )4 (0.250 m 2 )  5.00 10 6 C/m 2
After the introduction of
0.500  C
into the cavity, the outer charge is now
5.00  C  0.500 C  4.50 C
6
The surface charge density is now   q  q 2  4.50 10 C2  5.73 106 C/m2
A
4 r
4 (0.250 m)
(b) SET UP: Using Gauss’s law, the electric field is
E
E
q
q


A P0 A P0 4 r 2
EXECUTE: Substituting numbers gives
E
(8.85 1012
4.50 106 C
 6.47 105 N/C.
C2 /N  m2 )(4 )(0.250 m) 2
(c) SET UP: We use Gauss’s law again to find the flux.
E 
q
P0
.
EXECUTE: Substituting numbers gives
E 
0.500  106 C
 5.65  104 N  m2 /C2 .
8.85  1012 C2 /N  m2
EVALUATE: The excess charge on the conductor is still 5.00 C, as it originally was. The
introduction of the 0.500 C inside the cavity merely induced equal but opposite charges
(for a net of zero) on the surfaces of the conductor.
22.30. IDENTIFY: The net electric field is the vector sum of the fields due to each of the four
sheets of charge.
SET UP: The electric field of a large sheet of charge is
from a positive sheet and toward a negative sheet.
E   / 2P0 .
The field is directed away
EXECUTE:
(a) At A : EA 
EA 
1
(5 μ C m 2  2 μ C m 2  4 μ C m 2  6 μ C m 2 )  2.82 105 N C to the left.
2P0
(b) EB 
EB 
σ1
σ
σ
σ
1
 3  4  2 
( σ1  σ3  σ 4  σ 2 ) .
2P0 2P0 2P0 2P0 2P0
1
(6 μ C m 2  2 μ C m 2  4 μ C m 2  5 μ C m 2 )  3.95  105 N C to the left.
2P0
(c) EC 
EC 
σ2
σ
σ
σ
1
 3  4  1 
( σ 2  σ3  σ 4  σ1 ) .
2P0 2P0 2P0 2P0 2P0
σ4
σ
σ
σ
1
 1  2  3 
( σ 2  σ3  σ 4  σ1 ) .
2P0 2P0 2P0 2P0 2P0
1
(4 μ C m 2  6 μ C m 2  5 μ C m 2  2 μ C m 2 )  1.69  105 N C to the left
2P0
EVALUATE: The field at C is not zero. The pieces of plastic are not conductors.
22.44. IDENTIFY: Apply Gauss’s law and conservation of charge.
SET UP: Use a Gaussian surface that is a sphere of radius r and that has the point charge at
its center.
EXECUTE:
(a) For
ra, E
1 Q
,
4 P0 r 2
radially outward, since the charge enclosed is Q, the charge of the
point charge. For a  r  b ,
2Q
r b, E  1 2 ,
4πP0 r
E 0
since these points are within the conducting material. For
radially inward, since the total enclosed charge is –2Q.
(b) Since a Gaussian surface with radius r, for a  r  b , must enclose zero net charge, the
total charge on the inner surface is Q and the surface charge density on inner surface is
σ 
Q
4πa 2
.
(c) Since the net charge on the shell is 3Q and there is Q on the inner surface, there must
be 2Q on the outer surface. The surface charge density on the outer surface is   
2Q
.
4 b2
(d) The field lines and the locations of the charges are sketched in Figure 22.44a.
(e) The graph of E versus r is sketched in Figure 22.44b.
Figure 22.44
EVALUATE: For r  a the electric field is due solely to the point charge Q. For
electric field is due to the charge 2Q that is on the outer surface of the shell.
r b
the