11.3 Solutions
495
dE = dE z =
q sin θ dθ cos θ
4πε 0 a 2
∴
E =
dE z =
q
4πε 0 a 2
π/2
0
sin θ cos θ dθ =
q
8πε 0 a 2
(b) dV =
q sin θ dθ
4πε 0 a
; V =
dv =
q
4πε 0 a
π/2
0
sin θ dθ =
q
4πε 0 a
Fig. 11.29
11.28 The x-component of the field due to front charges will get cancelled and the
y-component is added up to
2q
4πε 0
a/2
r −
a
2
2 +
a 2
4
1/2
along the negative y-axis, Fig. 11.30.
Similarly the field due to the other two charges will be
2q · a/2
4πε 0
r +
a
2
2 +
a 2
4
3/2 along the positive y-axis.
Neglecting terms of the order of a 2 , the net field will be
E =
qa
4πε 0 r 3
1 −
a
r
−3/2 −
1 +
a
r
−3/2
Using the binomial expansion up to retaining terms linear in a,
E =
3qa 2
4πε 0 r 4 . Then E α
1
r 4
495
dE = dE z =
q sin θ dθ cos θ
4πε 0 a 2
∴
E =
dE z =
q
4πε 0 a 2
π/2
0
sin θ cos θ dθ =
q
8πε 0 a 2
(b) dV =
q sin θ dθ
4πε 0 a
; V =
dv =
q
4πε 0 a
π/2
0
sin θ dθ =
q
4πε 0 a
Fig. 11.29
11.28 The x-component of the field due to front charges will get cancelled and the
y-component is added up to
2q
4πε 0
a/2
r −
a
2
2 +
a 2
4
1/2
along the negative y-axis, Fig. 11.30.
Similarly the field due to the other two charges will be
2q · a/2
4πε 0
r +
a
2
2 +
a 2
4
3/2 along the positive y-axis.
Neglecting terms of the order of a 2 , the net field will be
E =
qa
4πε 0 r 3
1 −
a
r
−3/2 −
1 +
a
r
−3/2
Using the binomial expansion up to retaining terms linear in a,
E =
3qa 2
4πε 0 r 4 . Then E α
1
r 4