A parallel plate capacitor has plate `A` and separation `d` between the plates. The capacitor is connected to a battery of emf `V`. (a) find the charge on the capacitor. (b) the plate separation is decreases to `d//2`. Find the extra charge given by the battery to the positive plate. (c) work done in reducing separation between plates.

Explanation

(a) `C=(Ain_(0))/(d)`
`Q=CV=(Ain_(0)V)/(d)`
(b) `C'=(Ain_(0))/(d//2)=(2Ain_(0))/(d)`
`Q'=C'V=(2Ain_(0)V)/(d)`
Extra charge given by battery
`DeltaQ=Q'-Q=(Ain_(0)V)/(d)`
(c) `WS=U_(2)-U_(1)=(1)/(2)_C'V^(2)-(1)/(2)CV^(2)`
`=(V^(2))/(2)(C'-C)=(Ain_(0)V^(2))/(2d)`