502
11 Electrostatics
11.41 (a) q = C V = 4πε 0 r V =
5 × 10 6 × 10 −3
9 × 10 9
= 5.5 × 10
−7 C
(b) V =
q
4πε r
=
5.5 × 10 −7 × 9 × 10 9
1 × 10 −3
= 1.65 × 10
6 V
11.42 Initial potential energy U 1 = −
qe
4πε 0 r 1
Final potential energy U 2 = −
qe
4πε 0 r 2
U = U 1 − U 2 =
qe
4πε 0
1
r 2
−
1
r 1
By work–energy theorem, gain in kinetic energy = loss in potential energy.
1
2
mv
2
=
qe
4πε 0
1
r 2
−
1
r 1
v =
2qe
4πε 0 m
1
r 2
−
1
r 1
1/2
=
2 × 2 × 10 −9 × 1.6 × 10 −19 × 9 × 10 9
9.1 × 10 −31
1
0.18
−
1
1.2
1/2
= 5.467 × 10
6 m/s
11.43 V (r ) =
Q
4πε 0
2
r
−
1
√
r 2 + d 2
−
1
√
r 2 + d 2
=
2Q
4πε 0
1
r
−
1
r
1 +
d 2
r 2
−1/2
Qd 2
4πε 0 r 3
Thus V (r ) α
1
r 3
11.44 (i) F = q E = (1.6 × 10 −19 )(2 × 10 3 ) = 3.2 × 10 −16 N
(ii) Acceleration a =
F
m
=
3.2 × 10 −16
9.1 × 10 −31 = 3.516 × 10
14 m/s
2
v =
√
2as =
2 × 3.516 × 10 14 × 0.015
= 3.25 × 10
6 m/s
(iii) Outside the plates there is no force on the electron as there is no electric
field.
11.45 V (r ) =
Q
4πε 0 r
(1)
For continuous distribution of charge, each element dq can be treated as point
charge so that the contribution dV to the potential can be written according
to (1), Fig. 11.38: