Question

A parallel plate capacitor with air between the plates has a capacitance of 4 pF. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant 6, then the value of capacitance will be ……..

Hint:

The capacitance of a parallel plate capacitor increases by a factor of \( K \) when a dielectric is inserted. Reducing the plate distance further increases capacitance, as \( C \propto \frac{1}{d} \).

Answer 1

Step 1: Understanding the Capacitance Formula
The capacitance of a parallel plate capacitor is given by: \[ C = \frac{\varepsilon_0 A}{d} \] where:
- \( C \) is the capacitance,
- \( \varepsilon_0 \) is the permittivity of free space,
- \( A \) is the plate area,
- \( d \) is the separation between plates. When a dielectric of constant \( K \) is introduced, the capacitance becomes: \[ C' = K \frac{\varepsilon_0 A}{d} \] Step 2: Effect of Given Changes
- Given initial capacitance: \( C = 4 \) pF. - The plate separation is reduced by half, so new separation \( d' = \frac{d}{2} \). - A dielectric with constant \( K = 6 \) is introduced. The new capacitance is given by: \[ C' = K \frac{\varepsilon_0 A}{d'} \] Since \( d' = \frac{d}{2} \), we substitute: \[ C' = 6 \times \frac{\varepsilon_0 A}{(d/2)} \] \[ C' = 6 \times 2 \times \frac{\varepsilon_0 A}{d} \] \[ C' = 12C \] Step 3: Final Calculation
\[ C' = 12 \times 4 = 48 { pF} \] Thus, the new capacitance is 48 pF.