A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between the plates is filled with a dielectric of constant κ = 2. What is the relationship between the stored energies of the capacitor, where PEi is the initial stored energy, and PEf is the stored energy after the dielectric is inserted?

a. PEf = 2PEi

b. PEf = 0.25PEi

c. PEf = 0.5PEi

d. PEf = PEi

The relationship between the initial stored energy and the stored energy after the dielectric is inserted is:

c)

Explanation:

A parallel plate capacitor with that is connected to a voltage source holds a charge of . Then we disconnect the voltage source and keep the charge constant . If we insert a dielectric of between the plates while we keep the charge constant, we found that the potential decreases as:

The capacitance is modified as:

The stored energy without the dielectric is

The stored energy after the dielectric is inserted is:

If we replace in the above equation the values of V and C we get that

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