Question

A slab of material of dielectric constant K has the same area A as the plates of a parallel plate capacitor, and has thickness $\left( \frac{3}{4}d \right),$ where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is

• A. $C = \frac{\varepsilon_{0}A}{d}\left( \frac{K + 3}{4K} \right)$
• B. $C = \frac{\varepsilon_{0}A}{d}\left( \frac{2K}{K + 3} \right)$
• C. $C = \frac{\varepsilon_{0}A}{d}\left( \frac{2K}{K + 3} \right)$
• D. $C = \frac{\varepsilon_{0}A}{d}\left( \frac{4K}{K + 3} \right)$

Answer 1

Answer: (D)

$C = \frac{\varepsilon_{0}A}{d}\left( \frac{4K}{K + 3} \right)$

Answer 2

: Here, thickness of the slab,

$t = \frac{3}{4}d$ capacitance,

$C = \frac{\varepsilon_{o}A}{d - t\left( 1 - \frac{1}{K} \right)} = \frac{\varepsilon_{o}A}{d - \frac{3}{4}d\left( 1 - \frac{1}{K} \right)}$

$= \frac{\varepsilon_{o}A}{\frac{d}{4} + \frac{3}{4}\frac{d}{K}} = \frac{\varepsilon_{o}A}{\frac{d}{4}\left( 1 + \frac{3}{K} \right)} = \frac{\varepsilon_{o}A}{d}\frac{4K}{(K + 3)}$