Question

a. What is the potential difference between points a and 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 $b$ when switch $S$ is closed?
d. How much does the charge on each capacitor change when $S$ is closed?

Answer 1

i. In steady state, no current will flow through the circuit if $S$ is opened. So the potential at $a$ will be 18 V, and that at $b$ will be zero. Hence, $V_a-V_b=18-0=18V$.

ii. Obviously, $a$ is at the higher potential.

iii. When $S$ is closed, finally in steady state, current $I$ will flow as shown in Figure.

$I=\frac{18}{6+3}=2A,V_1=6I=6\times 2=12V$

$V_b=18-V_1=18-12=6V$

iv. $q_1=C_1{V}_1=6\times 12=72\mu C$

$V_1+V_2=18$ or $V_2=6V,q_2=C_2{V}_2=3\times 6=18,\mu C$

Before closing $S$, the potential on each capacitor is 18 V and charge is $q_{10}=6\times 18=108\mu C$, $q_{20}=3\times 18=54\mu C$

Change in charge is $q_1-q_{10}=-36\mu C$ and $q_2-q_{20}=-36\mu C$