11.3 Solutions
529
The potential drop across C 1 plus that across C 4 must be equal to the
potential drop across C 2 plus that across C 3 .
V 1 + V 4 = V 2 + V 3 = V
(1)
∴
q 1
C 1
+
q 1 − q 2
C 4
=
q − q 1
C 2
+
q − q 1 + q 2
C 3
(2)
By problem C 1 = C 2 = C 3 = C 4 .
∴ 2q 1 − q 2 = C 1 V
(3)
2q − 2q 1 + q 2 = C 1 V
(4)
Adding (3) and (4)
2q = 2C 1 V → C =
q
V
= C 1 = 2 μF
11.93 Applying the loop theorem to the circuit, traversing clockwise from the negative terminal of the battery we have the equation
ξ − i R −
q
C
= 0
( 1 )
where ξ is the emf of the battery and the second and the third terms represent
the potential drop across the resistor and the capacitor.
Now
i =
dq
dt
(2)
Using (2) in (1)
R
dq
dt
+
q
C
= ξ
(3)
This differential equation describes the time variation of the charge on the
capacitor. Re-arranging (3)
dq
Cξ − q
=
dt
RC
(4)
Integrating (4)
− ln (Cξ − q) =
t
RC
+ A
(5)
where A is the constant of integration which can be determined from the
initial condition.
At t = 0, q = 0 since the capacitor was uncharged.
∴
A = − ln ξ C
(6)