11.3 Solutions
483
(ii) q = −8 μC
F q = q E = −2.4 × 10
−3 N
Tension in the thread
T = 5.88 × 10
−3
− 2.4 × 10
−3
= 3.48 × 10
−3 N
(b) As the electric field is in the negative x-direction, point b will be at a
higher potential than a.
(i) Therefore V b − V a will be positive
(ii) E =
V b − V a
d
=
10 5
(6 − 2)
= 2.5 × 10 4 N/C
11.3 The total charge on the circular loop Q = 2π Rλ; the distance of the point
P(0, 0, Z ) from the loop is r = (Z 2 + R 2 ) 1/2 . Therefore, the electric potential
P will be
V =
Q
4πε 0 r
=
λR
2ε 0 (Z 2 + R 2 ) 1/2
E = −
∂ V
∂z
=
λR Z
2ε 0 (Z 2 + R 2 ) 3/2
11.4 (a) F =
q
4πε 0 r 2 (Q = 1)
V = −
F dr = −
q
4πε 0
dr
r 2 =
q
4πε 0 r
+ C
When r = ∞, V = 0. Therefore C = 0
(b) U =
q 2
4πε 0
1
d
−
1
d
+
1
d
−
1
d
−
√
2
d
−
√
2
d
For six pairs of charges
U = −
q 2
√
2πε 0 d
The potential energy does not depend on the order in which the charges
are assembled.
(c) Consider the forces on the top left-hand charge due to the three other
charges.
E x = E y = −
q 2
8 πε 0 d 2
∴ E =
E 2
x + E 2
y =
√
2
q 2
8 πε 0 d 2 = 0