Example

Question:

A slab of material of dielectric constant \(K\) has the same area as the plates of a parallel-plate capacitor but has a thickness \((3/4)d\), where \(d\) is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

Solution:

Let \(E_0 = V_0/d\) be the electric field between the plates when there is no dielectric and the potential difference is \(V_0\). If the dielectric is now inserted, the electric field in the dielectric will be \(E = E_0/K\).
The potential difference will then be: \[ V = E_0 \left( \frac{1}{4} d \right) + \frac{E_0}{K} \left( \frac{3}{4} d \right) = E_0 d \left( \frac{1}{4} + \frac{3}{4K} \right) = V_0 \frac{K + 3}{4K} \] The potential difference decreases by the factor \(\frac{K+3}{4K}\) while the free charge \(Q_0\) on the plates remains unchanged.
The capacitance thus increases: \[ C = \frac{Q_0}{V} = \frac{4K}{K+3} \frac{Q_0}{V_0} = \frac{4K}{K+3} C_0 \]