Question

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (which is connected across the capacitor terminals). How much energy is dissipated through the resistor in terms of U and K?

Answer 1

Answer: U/K

Explanation: In order to explain this problem we have to consider the energy stored in a capacitor which is giving by:

U=0.5*C*V^2 where C and V are the capacitance and the voltage, respectively.

When the capacitor is charged a dielectric material with dielectric constant K is inserted into the capacitor, filling the space between the plates.

It is well known then stored energy is decreased a factor K, so the final stored energy in the capacitor with dielectric is:

Ufinal=U/K

Thsi can justified by the following reasons:

the capacitance after insert the dielectric is given by:

C=KεoA/d  where A and d are the area and separaction of the plates of the capacitor. εo is a constant.

Cfinal=K*Cinitial

We also know that C=Q/V

as the charge is constant in the capacitor,  then the voltage is decreased  a factor K, then the energy is:

Ufinal=0.5*Cfinal*Vfinal^2=0.5*K*Cinitial*Vinitial^2/(K^2)=0.5**Cinitial*Vinitial^2/K=U/K

Finally this energy can be dissipated through the resistor.