Question

A parallel plate capacitor, in which the space between the plates is filled with a dielectric material with dielectric constant κ = 9.8, has a capacitance of C=472 μF and it is connected to a battery whose voltage is V=30 V and fully charged. Once fully charged, while still connected to the battery, dielectric material is removed from the capacitor.

How much change occurs in the energy of the capacitor; final energy minus initial energy?

Answer 1

Given that Initial capacitance = C0 = 472*10^-6 F

V0 = Potential across battery = 30 V

So Initial charge on capacitor = Q0 = C0*V0 = 472*10^-6*30 = 14.16*10^-3 C

Now energy stored in capacitor will be:

U0 = Q0^2/(2*C0) = (14.16*10^-3)^2/(2*472*10^-6) = 212.4*10^-3 J = 212.4 mJ

Now when battery is disconnected then charge will remain constant, So

Q1 = Q0 = 14.16*10^-3 C

But new capacitance will be:

C1 = C0/k = 472*10^-6/9.8 = 48.16326*10^-6 C

So final energy stored will be:

U1 = Q1^2/(2*C1) = (14.16*10^-3)^2/(2*48.16326*10^-6) = 2081.52*10^-3 J = 2081.52 mJ

So change in potential energy will be:

dU = U1 - U0

dU = 2081.52 - 212.4 = 1869.12 mJ

dU = 1869 mJ