Question

Consider the situation shown in figure. The switch $S$ is open for a long time and then closed. Find the charge flown through the battery when the switch is closed.

• A. $CE/2$
• B. $CE$
• C. $CE/4$
• D. $0$

Answer 1

The correct option is A $CE/2$

When $S$ is open. Both capacitors are in series and its equivalent capacitance will be

$C_{eq}=\frac{C}{2}$
Charge on the each plate of capacitor will be

$Q=\frac{CE}{2}$

So charge distribution on the plate of PPC will be as shown below

When $S$ is closed: $\Delta V$ across the capacitor on the right is $0$.
So,
$Q_f=CE$


Thus, charge flown through battery will be

$\Delta Q=CE-\frac{CE}{2}=\frac{CE}{2}$

Hence, option $\left(a\right)$ is correct.