AP Physics 2 Exam Question

Consider a parallel plate capacitor with plate area $A$ and plate separation $d$. The capacitor is connected to a battery of voltage $V_b$. The dielectric material between the plates has a dielectric constant $\epsilon_r$.

a) Calculate the capacitance $C$ of the capacitor.

b) If the battery is disconnected and a dielectric slab with thickness $l$ is inserted between the plates, calculate the new capacitance $C'$.

c) Explain how the insertion of a dielectric slab affects the electric field between the plates.

Exam Question Solution

a) The capacitance of a parallel plate capacitor is given by the formula:

where $\epsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2$. Therefore, the capacitance $C$ is:

b) When a dielectric slab is inserted between the plates of a parallel plate capacitor, its capacitance increases by a factor equal to the dielectric constant $\epsilon_r$ of the slab. Therefore, the new capacitance $C'$ is given by:

where $l$ is the thickness of the dielectric slab.

c) The insertion of a dielectric reduces the effective electric field between the plates of the capacitor. This occurs because the electric field lines passing through the dielectric material experience a "slowing down" or reduction in strength due to the polarization of the dielectric. The dielectric polarizes, creating regions of positive and negative charge, which reduces the overall electric field between the plates.

Therefore, the electric field between the plates is weakened by a factor equal to the relative permittivity $\epsilon_r$ of the dielectric material.