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 given by: \[ C = \frac{\epsilon_0 A}{d} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates. When a dielectric is inserted between the plates, the capacitance increases by a factor of \( K \), the dielectric constant. The cases can be analyzed as follows: Case (a) In this case, the capacitor is filled with the dielectric constant \( K \) in half of the space between the plates. We treat the capacitor as two capacitors in series, one filled with the dielectric and the other without it.
- The first capacitor has a dielectric material with constant \( K \) and a distance of \( \frac{d}{2} \).
- The second capacitor has no dielectric material and also has a distance of \( \frac{d}{2} \). 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: \[ C_2 = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d} \] Since the capacitors are in series, the total capacitance is: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting the values of \( C_1 \) and \( C_2 \): \[ \frac{1}{C} = \frac{1}{\frac{2 K \epsilon_0 A}{d}} + \frac{1}{\frac{2 \epsilon_0 A}{d}} \] Simplifying: \[ \frac{1}{C} = \frac{d}{2 K \epsilon_0 A} + \frac{d}{2 \epsilon_0 A} \] \[ \frac{1}{C} = \frac{d}{2 \epsilon_0 A} \left( \frac{1}{K} + 1 \right) \] Thus, the total capacitance is: \[ C = \frac{2 \epsilon_0 A}{d} \left( \frac{K}{K+1} \right) \] Case (b) In this case, the entire space between the plates is filled with the dielectric material with dielectric constant \( K \). The capacitance is simply: \[ C = \frac{K \epsilon_0 A}{d} \]