Answers to Electric Field Homework
21.Use Eq. 16–4a to calculate the electric field due to a point charge.
E=k
Q
r
2
= (8.988 ´ 109 N × m 2 /C2 )
33.0 ´ 10-6 C
(0.217 m)
2
= 6.30 ´ 106 N/C upward
28.The field due to the negative charge will point toward the negative charge, and the field due to the
positive charge will point toward the negative charge. Thus the magnitudes of the two fields can be
added together to find the charges.
Enet = 2EQ = 2k
Q
( /2)2
=
8kQ
2
® Q=
E 2
(386 N/C)(0.160 m)2
=
= 1.37 ´ 10-10 C
8k 8(8.988 ´ 109 N × m 2 /C2 )
32.For the net field to be zero at point P, the magnitudes of the fields created by Q1 and Q2 must be equal.
Also, the distance x will be taken as positive to the left of Q1. That is the only region where the
total field due to the two charges can be zero. Let the variable
Q
Q2
E1 = E2 ® k 21 = k
x
(x + )2
x=
(
Q1
Q2 - Q1
)
= (12 cm)
(
represent the 12-cm distance.
®
32 mC
45mC - 32 mC
)
= 64.57 cm » 65 cm
36.We assume that gravity can be ignored, which is proven in part (b).
(a)
The electron will accelerate to the right. The magnitude of the acceleration can be found from
setting the net force equal to the electric force on the electron. The acceleration is constant, so
constant-acceleration relationships can be used.
Fnet = ma = |q|E ® a =
|q| E
m
u 2 = u02 + 2aDx ® u = 2aDx = 2
= 2
|q|E
Dx
m
(1.602 ´ 10-19 C)(1.45 ´ 104 N/C)
(9.11 ´ 10-31 kg)
(1.60 ´ 10-2 m) = 9.03 ´ 106 m/s