Question

Two identical parallel plate capacitors A and B connected to a battery of V volts w/ the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled w/ a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer 1

The electrostatic energy stored in a capacitor with capacitance
`C_0`, with a voltage difference V applied to it, and without dielectric, is given by

`U_0 = (1)/(2) C_0 V^2`
Now let's assume we fill the space between the two plates of the capacitor with a dielectric with constant k. The new capacitance of the capacitor is

`C_k = k C_0`
So, the energy stored now is

`U_k = (1)/(2)C_k V^2= (1)/(2)kC_0 V^2`

Therefore, the ratio between the energies stored in the capacitor before and after the introduction of the dielectric is

`(U_k)/(U_0)= ( (1)/(2)kC_0 V^2 )/( (1)/(2) C_0 V^2)= k`