Question

A parallel plate capacitor, in which the space between the plates is filled with a dielectric material with dielectric constant κ = 6.2, has a capacitance of C=4,650 μF and it is connected to a battery whose voltage is V=37 V and fully charged. Once it is fully charged, it is disconnected from the battery and without affecting the charge on the plates, dielectric material is removed from the capacitor.

How much change occurs in the energy of the capacitor; final energy minus initial energy? Express your answer in units of mJ (milli Joules) using zero decimal places.

Answer 1

Given that Initial capacitance = C0 = 4650*10^-6 F

V0 = Potential across battery = 37 V

So Initial charge on capacitor = Q0 = C0*V0 = 4650*10^-6*37 = 172.05*10^-3 C

Now energy stored in capacitor will be:

U0 = Q0^2/(2*C0) = (172.05*10^-3)^2/(2*4650*10^-6) = 3.1829 J = 3182.9 mJ

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

Q1 = Q0 = 172.05*10^-3 C

But new capacitance will be:

C1 = C0/k = 4650*10^-6/6.2 = 750*10^-6 F

So final energy stored will be:

U1 = Q1^2/(2*C1) = (172.05*10^-3)^2/(2*750*10^-6) = 19.7341*10^-3 J = 19734.1 mJ

So change in potential energy will be:

dU = U1 - U0

dU = 19734.1 - 3182.9 = 16551.2 mJ

dU = 16551 mJ