A parallel plate capacitor has capacitance C. If the charges of the plates are `Q` and `-3Q`, find of the

i. charges at the inner surfaces of the plates
ii. Potential difference between the plates
iii. charge flown if the plates are connected
iv. energy lost by the capacitor in (iii.)
v. charge flown if any plate is earthed.

Explanation

The charge on surface (ii), `q_(2)=(Q-(-3Q))/2=2Q`
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i. Hence the charges on the inner surfaces are `2Q` and `-2Q`.
ii. `V=("Charge of capacitor")/("Capacitance")=(2Q)/C`
iii. If the plates are connected, `2Q` and `-2Q` will be neutralized, so `2Q` charge will flow the from the left to the right plate.
iv. Energy lost `=` initial energy stored in capacitor
`=(2Q)^(2)/(2C)=(2Q^(2))/C`
v. If the left plate is earthed, `2Q` charge flows from the ground to the left plate. If the right plate is earthed, `2Q` charge flows from the from the ground to right plate.
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