Question

A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases.

Hint:

When dielectric materials are inserted between the plates of a capacitor, the capacitance increases. For a parallel plate capacitor with different dielectric materials, we can use series and parallel combinations depending on the geometry of the dielectric placement.

Answer 1

For a parallel plate capacitor, the capacitance \( C \) is defined by the equation:\[C = \frac{\epsilon_0 A}{d}\]with the following parameters:
- \( \epsilon_0 \): permittivity of free space,
- \( A \): area of the capacitor plates,
- \( d \): separation distance between the plates.Introducing a dielectric material between the plates increases capacitance by a factor of \( K \), the dielectric constant. The scenarios are analyzed as follows: Case (a)This scenario involves filling half the inter-plate space with a dielectric material of constant \( K \). The capacitor is modeled as two series-connected capacitors: one containing the dielectric and the other being empty.
- Capacitor 1: dielectric material with constant \( K \), separation \( \frac{d}{2} \).
- Capacitor 2: no dielectric material, separation \( \frac{d}{2} \).The capacitance of the first capacitor is:\[C_1 = \frac{K \epsilon_0 A}{d/2} = \frac{2 K \epsilon_0 A}{d}\]The capacitance of the second capacitor is:\[C_2 = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d}\]For capacitors in series, the total capacitance \( C \) is determined by:\[\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}\]Substituting \( C_1 \) and \( C_2 \):\[\frac{1}{C} = \frac{1}{\frac{2 K \epsilon_0 A}{d}} + \frac{1}{\frac{2 \epsilon_0 A}{d}}\]After simplification:\[\frac{1}{C} = \frac{d}{2 K \epsilon_0 A} + \frac{d}{2 \epsilon_0 A}\]\[\frac{1}{C} = \frac{d}{2 \epsilon_0 A} \left( \frac{1}{K} + 1 \right)\]Consequently, the total capacitance is:\[C = \frac{2 \epsilon_0 A}{d} \left( \frac{K}{K+1} \right)\] Case (b)In this scenario, the entire space between the plates is filled with a dielectric material of constant \( K \). The capacitance is calculated directly as:\[C = \frac{K \epsilon_0 A}{d}\]