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Q. A parallel plate capacitor has capacitance $C$. If it is equally filled the parallel layers of materials of dielectric constant $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1 $ and $C$ is

VITEEEVITEEE 2012

Solution:

$C_{A} = \frac{K_{1} \varepsilon_{0}A}{\frac{d}{2}} , C_{B} = \frac{K_{2}\varepsilon_{0}A}{\frac{d}{2}} $
$C_{eq} = \frac{C_{1}}{C_{2}} = \frac{2K_{1}K_{2}}{K_{1} + K_{2}}$
$ = \frac{C_{A}C_{B}}{C_{A} + C_{B}} = \left(\frac{2K_{1} K_{2}}{K_{1} + K_{2}}\right) \frac{\varepsilon_{0}A}{d} $
$\left( \because e = \frac{\varepsilon_{0}A}{d}\right) $