Charges $Q_{1} a n d Q_{2}$ are given to two plates of a parallel plate capacitor. The capacity of the capacitor is C. When the switch is closed, mark the correct statement(s). (Assume both $Q_{1} a n d Q_{2}$ to be positive)

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The charge flowing through the switch is zero

The charge flowing through the switch is $Q_{1} + Q_{2} .$

Potential difference across the capacitor plate is $Q_{1} / C .$

The charge of the capacitor is $Q_{1}$

answer is B, C, D.

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Detailed Solution

As the switch is closed, the charge on the outer surface of the second plate becomes zero. From the concept in electrostatics that electric field inside the bulk of the material of conductor is zero, we can find the charges on various faces. So it is clear that $Q_{1} + Q_{2}$ charge goes from the second plate to the earth. Charge on capacitor is $Q_{1}$ and hence its potential is $Q_{1} / C .$