526
11 Electrostatics
(ii) At one time constant t = RC
V = V 0 e
−1
=
120
2.718
= 44.15 V
U =
1
2
C V
2
=
1
2
× 10 × 10
−6
× (44.15)
2
= 0.097 J
(iii) H = i 2 Rt =
V 2
R
t = V 2 C = (44.15) 2 × 10 × 10 −6 = 0.0195 J
11.88 Equilibrium energy U 0 =
1
2
C V 2
0
Energy at time t
U =
1
2
C V
2
=
1
2
C V
2
0
1 − e
−t/RC
2 = U 0
1 − e
−t/RC
2
U
U 0
=
1
2
=
1 − e
−t/RC
2
Solving t = 1.228RC.
Thus after 1.228 time constants the energy stored in the capacitor will reach
half of its equilibrium value.
11.89 Let the capacitor be divided into differential strips which are practically parallel. Consider a strip at distance x of length a perpendicular to the plane
of paper and of width dx in the plane of paper, the area of the strip being
dA = adx, Fig. 11.46. At the distance x, the separation of the plates is seen
to be t = D + xθ . The capacitance due to the differential strip facing each
plate is
dC =
ε 0 dA
D
=
ε 0 adx
D + x θ
Fig. 11.46