484
11 Electrostatics
Therefore the charge is not in equilibrium. Same thing is true for the other
three charges.
11.5 (i) V 1 =
q
4πε 0 r
=
2 × 10 −15 × 9 × 10 9
10 −3
= 0.018 V
(ii) If n droplets each of radius r coalesce to form a large drop of radius R,
then assuming that the droplets are incompressible, the volume does not
change. Therefore
4
3
π R
3
= n
4
3
πr
3
or
R = n
1/3 r
(1)
As charge is conserved
Q = nq
(2)
where Q and q are the charges on the drop and droplet, respectively. Then
the potential of the droplet is
V 2 =
Q
4πε 0 R
=
nq
4πε 0 n 1/3 r
=
n 2/3 q
4πε 0 r
= n
2/3 V 1 = 2
2/3 V 1 = 0.0286 V
where n = 2, and V 1 = 0.018 V by (i).
11.6 Electric force
F = q E = 2 × 10
−8
× 20, 000 = 4 × 10
−4
Gravitational force mg = 80 × 10 −6 × 9.8 = 7.84 × 10 −4 N
Balancing the horizontal and vertical components of forces, Fig. 11.19
T sin θ = F
(1)
T cos θ = mg
(2)
where T is the tension in the thread.
Fig. 11.19