Question

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

Answer 1

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$. Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     $V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

                                              $C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

                                               $PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

                                               $PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

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

                                         $PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

                                                   $PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

                                                  $PE_{f} =0.5\ PE_{i}$