Question

What is the effect on the capacitance of a parallel plate capacitor when
(i) distance between the plates is doubled.
(ii) area of the plates is halved.
(iii) a dielectric medium is filled between the plates.

Hint:

Remember the dependencies for a parallel plate capacitor: Capacitance is directly proportional to the plate area (A) and the dielectric constant (K), and inversely proportional to the distance between the plates (d).

Answer 1

Step 1: Key Formula:
The capacitance (C) of a parallel plate capacitor with plate area A, distance between plates d, and vacuum (or air) between the plates is given by the formula: \[ C = \frac{\epsilon_0 A}{d} \] where \(\epsilon_0\) is the permittivity of free space. We will use this formula as the baseline for all three cases.
Step 2: Analysis of each case:
(i) Distance between the plates is doubled.
The initial capacitance is \( C = \frac{\epsilon_0 A}{d} \).
The new distance is \( d' = 2d \).
The new capacitance, \( C' \), will be: \[ C' = \frac{\epsilon_0 A}{d'} = \frac{\epsilon_0 A}{2d} = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) = \frac{1}{2} C \] Effect: The capacitance is halved.
(ii) Area of the plates is halved.
The initial capacitance is \( C = \frac{\epsilon_0 A}{d} \).
The new area is \( A' = A/2 \).
The new capacitance, \( C' \), will be: \[ C' = \frac{\epsilon_0 A'}{d} = \frac{\epsilon_0 (A/2)}{d} = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) = \frac{1}{2} C \] Effect: The capacitance is halved.
(iii) A dielectric medium is filled between the plates.
When a dielectric medium of dielectric constant K is completely filled between the plates, the capacitance formula becomes: \[ C' = \frac{K \epsilon_0 A}{d} \] Since the original capacitance was \( C = \frac{\epsilon_0 A}{d} \), we can write: \[ C' = K . C \] For any dielectric medium, the dielectric constant K is always greater than 1 (\(K>1\)).
Effect: The capacitance increases by a factor of K.