Question

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch $S_1$ is pressed first to fully charge the capacitor $C_1$ and then released. The switch $S_2$ is then pressed to charge the capacitor $C_2$. After some time, $S_2$ is released and then $S_3$ is pressed. After some time :

• A. the charge on the upper plate of $C_1$ is $2C{V}_0$
• B. the charge on the upper plate of $C_1$ is $C{V}_0$
• C. the charge on the upper plate of $C_2$ is 0
• D. the charge on the upper plate of $C_2$ is $-C{V}_0$

Answer 1

Answer: (B), (D)

When $S_1$ is pressed, the capacitor $C_1$ will start to charge and its upper plate get charge $+2C{V}_0$ and lower plate $-2C{V}_0$.
When $S_1$ is opened and $S_2$ pressed , the total charge of system will distribute equally into $C_1$ and $C_2$ because their capacitance is same. Thus, upper plate of each capacitor will get charge $+C{V}_0$ and lower plate $-C{V}_0$
Ans :B