A capacitor C is connected to a battery of V volts and is fully charged.Keeping

Question

A capacitor C is connected to a battery of V volts and is fully charged.Keeping the battery connected, the spacing between the capacitor plates is reduced to half.

What happens to the potential difference between the two plates? Why?

What happens to the charge on the capacitor? Why?

What happens to the energy stored in the capacitor? Why?

Now the battery is disconnected, and then the plate spacing is restored to its original value.

What happens to the potential difference between the two plates? Why?   

What happens to the charge on the capacitor? Why?

What happens to the energy stored in the capacitor? Why?

Note: It is not enough if you say

Explanation / Answer

use the formulas

Capaciatnce C = Q/V     or   C0 = eoA/d

energy U = 0.5 CV^2   or   0.5 QV

also Vo = Ed

so

when d is reduced to half ,

C = eoA/d/2   = 2 Co

and V = Ed/2

so PD reduces by half
------------------------

as Q = CV,   as V reduces by 2,

Q also reduces by a factor of 2

-------------------------

Eenrgy U = 0.5 * Q * V^2/4

U reduces by a factor of 2

--------------------------------------

V will eqaul to initial Volatge

Q charge will remains same as Original

Eenrgy remains same as Original

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