A parallel plate capacitor of capacitance C0 has plates of area A with separation d between them. When it is connected to a battery of voltage V0, it has charge of magnitude Q0 on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are

Question

Quizplus · Accepted Answer

When a dielectric is added between the plates, the capacitance increases by a factor of the dielectric constant, i.e. C = C0*k, where k = 3. Therefore, Q = CV = C0*k*V0. Since the battery is disconnected, the charge on the plates remains constant, i.e. Q = Q0. Therefore, V = Q/C = Q0/(C0*k) = Q0/C, which is equal to V0/3. So, the correct answer is option B.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area