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. (a) The charge on the upper plate of `C_1` is `2CV_0`
B. (b) The charge on the upper plate of `C_1` is `CV_0`
C. (c) The charge on the upper plate of `C_2` is 0
D. (d) The charge on the upper plate of `C_2` is `-CV_0`

Answer

B, D

Explanation

Step1: When `S_1` is pressed. The capacitor `C_1` gets charged such that its upper plate acquires a positive charge `+2CV_0` and lower plate `-2CV_0`.
Step 2: When `S_2` is pressed (`S_1` open). As `C_1=C_2` the charge gets distributed equal. The upper plates of `C_1` and `C_2` now take charge `+CV_0` each and lower plate `-CV_0` each.
(b) and (d) are correct option.