504
11 Electrostatics
(c) If x >> R, (6) can be expanded binomially,
V →
σ
2ε 0
x
1 +
R 2
x 2
1/2
− x
=
σ R 2
4ε 0 x
=
q
4πε 0 x
, an expression
which is appropriate for the point charge. This result is reasonable since
at very large distances the disc appears as a point.
11.46 For circular motion of electron, the speed
v =
2πr
T
(1)
For a stable orbit, the centripetal force = electric force.
mv 2
r
=
1
4πε 0
e 2
r 2
(2)
Eliminating v between (1) and (2)
T
2
= 16π
3
ε 0 m r
3
or T
2
α r
3
(Kepler’s third law)
11.47 Figure 11.39 shows the forces on charge 2 due to the charges 1, 3 and 4. The
forces
F 12 = F 32 =
1
4πε 0
Q 2
a 2
F 42 =
1
4πε 0
Q 2
√
2a
2 =
1
4πε 0
Q 2
(2a 2 )
Fig. 11.39
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