(a) What is the potential of point a with respect to point b in figure, when switch `S` is open?
(b) Which point, `a` or `b`, is at the higher potential?
(c) What is the final potential of point `b` with respect to ground when switch S is closed?
(d) How much charge flows through switch `S` when it is closed?

Explanation

i. In steady state, no current will
flow throgh the circuit is S
is opened. So the potential at
a will be 18V, and that at b
will be zero. Hence, `V_(alpha) -V_b`
=18 - 0 = 18V.
ii. Obviously, a is at the higher
potential.

iii. When S is closed, finally in
steady state, current I will
flow as shwon in fig.
`I = (18)/(6+3) = 2A, V_1 =6I = 6xx2 = 12V`
`V_b = 18 - V_1 = 18-12= 6V`
iv. `q_1 = C_1V_1 = 6xx12 =72muC`
`V_1 + V_2 = 18 or V_2 = 6V, q2 = C_2V_2 = 3xx6 = 18muC`
Before closing S, the potentail on each capacitor is 18 V
and charge is `q_(10) = 6xx18 = 108muC, q_(20) = 3xx18 - 54muC`
Change in charge is `q_1 - q_(10) = -36muC and q_2 - q_(20) = -36muC`