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 capacitance and the potential difference between the plates are

Question

Quizplus · Accepted Answer

The capacitance of a parallel-plate capacitor with a dielectric between its plates is given by $C = \kappa \epsilon_0 A/d$, where $\kappa$ is the dielectric constant, $\epsilon_0$ is the vacuum permittivity, $A$ is the area of the plates, and $d$ is the separation between the plates.

After the dielectric is added, the capacitance becomes $C' = 3C_0$.

Since the battery is ideal, the potential difference between the plates remains the same before and after the dielectric is added, i.e., $V' = V_0$.

Thus, the correct answer is (D) 3C0, 11eb8c7e_f590_6189_8371_e97dcabfcd5a_TB8847_11 V0.

A Parallel Plate Capacitor of Capacitance C0 Has Plates of Area