Question

(i) A dielectric slab of dielectric constant \( K \) and thickness \( t \) is inserted between plates of a parallel plate capacitor of plate separation \( d \) and plate area \( A \). Obtain an expression for its capacitance.

Hint:

A dielectric increases the capacitance of a capacitor, but if it doesn't fully fill the gap, the system behaves as capacitors in series.

Answer 1

Capacitance with a Dielectric Slab
- The capacitance of a parallel plate capacitor in vacuum is: \[ C_0 = \frac{\epsilon_0 A}{d} \] where:
- \( \epsilon_0 \) = Permittivity of free space,
- \( A \) = Plate area,
- \( d \) = Plate separation.
- When a dielectric slab of thickness \( t \) and dielectric constant \( K \) is inserted:
- The remaining air gap in the capacitor is \( (d - t) \). - The system now behaves as two capacitors in series:
1. Capacitor with dielectric: \( C_1 = \frac{K \epsilon_0 A}{t} \).
2. Capacitor with air gap: \( C_2 = \frac{\epsilon_0 A}{d - t} \). - The effective capacitance is: \[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ \frac{1}{C} = \frac{t}{K \epsilon_0 A} + \frac{(d - t)}{\epsilon_0 A} \] \[ C = \frac{\epsilon_0 A}{\frac{t}{K} + (d - t)} \] Thus, the capacitance of the capacitor with a dielectric slab is: \[ C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}} \]