Suppose a parallel plate capacitor (with capacitance C_o) is fully charged (to a

Question

Suppose a parallel plate capacitor (with capacitance C_o) is fully charged (to a value Q_o) by a battery. The battery (which supplies a potential difference of V_o) stays connected to the capacitor. If the plates of the capacitor are then moved closer together (the separation distance d between the plates is halved), describe quantitatively what happens to: the capacitance of the capacitor. the potential difference between the plates. the energy stored in the capacitor. the charge on the plates.

Explanation / Answer

we know
Co = A*epsilon/d

Qo = Co*Vo

Uo = (1/2)*Co*V^2

a) C' = A*epsilon/(d/2)

= 2*A*epsilon/d

= 2*Co

so, capaciatnce becomes doubled.

b) The potential difference between the plates stays the same.

V' = Vo

c) U' = (1/2)*C'*V'^2

= (1/2)*2*Co*Vo^2

= 2*Uo

so, The energy stored becomes double

d) Q' = C'*V'

= 2*Co*Vo

= 2*Qo

so, the charge on the capacitor plates becomes double.

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