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

Options

A. the charge on the upper plate of `C_1 is 2CV_0`
B. the charge on the upper plate of `C_1 is CV_0`
C. the charge on the upper plate of `C_1 is 0`
D. the charge on the upper plate of `C_2 is -CV_0`

Explanation

After switch `S_(1)` is closed, `C_(1)` is charged by `2CV_(0)`, when switch `S_(2)` is closed `C_(1)` and `C_(2)` both have plate charge `CV_(0)`. When `S_(3)` is closed, then upper plate of `C_(2)` becomes charged by `- CV_(0)` and lower plate by `+ CV_(0)`