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

Answer 1

voltage is constant since the battery is connected

Since capacitance C is given by

C = K * A / d

if we reduce the distance between the plates by half the capacitance will be doubled

that is new capacitance = K * A / (d/2)

new capacitance = 2 * K * A / d

new capacitance = 2 * C

charge on capacitor = C * V

new by reducing the distance the capacitance is doubled so

new charge on capacitor = new capacitance * V

new charge on capacitor = 2 * C * V

new charge on capacitor = 2 * charge on capacitor

since energy is directly proportional to capacitance that is 0.5 * C * V^2

so potential energy will also get doubled

if we remove the battery then restore the distance between plates then

the capacitance will change to C from 2 * C as distance will change from d/2 to d

since the energy will remain conserved so

0.5 * 2 * C * Vold^2 = 0.5 * C * Vnew^2

so new potential difference = * V

since charge = C * V

and the voltage is changed to * V so

charge = C * * V

energy will remain same