Download Chapter 21 solutions

21.11: F2 is in the x-direction, so F1 must be in the x-direction and q1 is positive.
q q
qq
F1  F2 ,
k 12 3  k 2 2 3
r13
r23
q1  0.0200 0.0400 q2  0.750 nC
2
21.17: F2 is in the x-direction.
qq
F2  k 1 2 2  3.37 N, so F2 x   3.37 N
r12
Fx  F2x  F3 x and F x  7.00 N
F3x  Fx  F2 x  7.00 N  3.37 N  10.37 N
For F3x to be negative, q3 must be on the –x-axis.
F3  k
q1q3
, so x 
x2
kq1q3
 0.144 m, so x   0.144 m
F3
21.21: a)
qQ
1
2qQa
.
2 sin 
2
4 0 (a  x )
4 0 (a  x2 )3 2
1 2qQ
c) At x  0, F 
in the y direction.
4 0 a 2
b) Fx  0, Fy  2
d)
1
2
21.25: Let +x-direction be to the right. Find a x :
v0x  1.50  103 m s, vx  1.50  103 m s, t  2.65  10 6 s, a x  ?
vx  v0x  a xt gives a x  1.132  109 m s2
18
Fx  ma x  7.516  10
N
F is to the left x-direction , charge is positive, so E is to the left.
E  F q  (7.516  10 18 N)
2 (1.602  10
19

C)  23.5 N C
21.26: (a) x  12 at 2
2x
2(4.50 m)
2
a 2 
 1.00  1012 m s
6
2
t
(3.00  10 s)
2
F ma (9.11  10 31 kg) (1.00  1012 m s )
E 

q
q
1.6  10 19 C
 5.69 N C
The force is up, so the electric field must be downward since the electron is negative.
(b) The electron’s acceleration is ~1011 g, so gravity must be negligibly small
compared to the electrical force.
21.27: a) q E  mg  q 
(0.00145 kg) (9.8 m s2 )
 2.19  105 C, sign is negative.
650 N C
(1.67  10 27 kg) (9.8 m s2 )
1.02  107 N/C, upward.
b) qE  mg  E 
19
1.60  10 C
9
2
2
9
ˆ (9 10 N  m C ) (5.00  10 C)  (2.813 10 4 N C) ˆj
21.32: a) E1 
2 j 
4 0r1
0.0400 m2
q1
9
E2 
2
2
9
q2 (9  10 N m C ) (3.00 10 C)
4

1.080 10 N C
2
2
r22
0.0300m  (0.0400 m)
The angle of E2 , measured from the x-axis, is 180  tan 1

4.00 cm
3.00cm
 126.9 Thus
E2  (1.080  104 N C) ( ˆi cos 126.9  ˆj sin 126.9)
3
3
 (  6.485  10 N C) iˆ  (8.64  10 N C) ˆj
b) The resultant field is
E1  E2  (  6.485  103 N C) iˆ  (  2.813  104 N C  8.64  103 N C) ˆj
 (  6.485  103 N C) iˆ  (1.95  10 4 N C) ˆj
 
  1.35

12 
2ˆ
2 ˆ
1
  , rˆ   ˆj b) tan
21.37: a) tan 
 , rˆ 
i
j

0
2
.2
4
2
2
1
 2.6 
c) tan 1 
  1.97 radians 112.9, rˆ   0.39iˆ  0.92 ˆj (Second quadrant).
 1.10
21.50: For a ring of charge, the electric field is given by Eq. (21.8).
1
Qx
9
ˆ
C, a  0.025 m
a) E 
2
2 3 2 i so with Q  0.125  10
4 0 (x a )
and x  0.4 m  E  7.0iˆ N C.
6
5
b) Fon ri ng   Fon q   q E   (  2.50  10 C) (7.0iˆ N C)  1.75  10 iˆ N.
21.54:
a)
b)
c)
By superposition we can add the electric fields from two parallel sheets of charge.
E  0.
E  0.
E  2 2 0  0 , directed downward.