Question

A parallel plate capacitor with capacitance C0 is connected to a battery with voltage V0 and is charged fully to a charge Q0. There is an electric field E0 between the plates and energy U0 stored in the capacitor.

With the battery still connected, the area of the plates are doubles and the plates are pulled to a distance three times its original distance (while still being small compared to the size of the plates).

Find (in terms of the original variables):

The new voltage V

The new charge C

The new electric field E

The new energy stored U

Answer 1

C = epsilon₀*A/d

New capacitance, C' = epsilon₀*2A/3d = (2C0/3)

As the battery is still connected , voltage across the plates will not be changed

So a) The new voltage V' = V0 (ans)

b) Q = CV

New charge Q' = C'*V' = (2/3)*C*V0 = (2/3)Q0 (ans)

c) E = V/d

New electric field, E' = V'/d' = V0/3d = (E0/3) (ans)

d) U = 1/2*C*V²

C' =(2/3)C , V' =V0

So U'= 1/2*C'V'² = (2/3)*1/2*C0*V0² = (2/3)U0 (ans)