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_(2) is 0`
D. The charge on the upper plate of `C_(2) is -CV_(0)`

Answer

B, D

Explanation

After pressing `S_1` chare on upper plate `c_1` is `cv_0`.After pressing `S_2` this charge equally distributes in two capicators. Therefore charge on upper plates of both capacitor will be `+CV_0`. When `S_2` is released and `S_2`pressed,charge on upper plate of `c_1` remains unchanged `(=+cv_0)` but charge on upper plate of is according to new battery `(=-cv_0)`