Question

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant K is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is \(\frac{3}{4}d\), where 'd' is the separation between the plates of parallel plate capacitor. The new capacitance (C') in terms of original capacitance (\(C_0\)) is given by the following relation :

• A. \(C' = \frac{3 + K}{4K}\,C_0\)
• B. \(C' = \frac{4 + K}{3}\,C_0\)
• C. \(C' = \frac{4K}{K + 3}\,C_0\)
• D. \(C' = \frac{4}{3 + K}\,C_0\)

Hint:

With a slab of thickness \(t\) filling the plate area, \(C' = \frac{\epsilon_0 A}{d - t + t/K}\) with \(t = 3d/4\).

Answer 1

The Correct Option is C

Original capacitance: \(C_0 = \frac{\epsilon_0 A}{d}\).
With a dielectric slab of thickness \(t = \frac{3d}{4}\) and the same area, the effective separation is \(d - t + \frac{t}{K} = d - \frac{3d}{4} + \frac{3d}{4K} = \frac{d}{4} + \frac{3d}{4K} = \frac{d(K + 3)}{4K}\).
\(C' = \frac{\epsilon_0 A}{d}\cdot\frac{4K}{K + 3} = \frac{4K}{K + 3}\,C_0\).
Answer: (3) \(C' = \frac{4K}{K + 3}\,C_0\)