Q. A parallel plate capacitor has the space between its plates filled by two slabs of thickness $\frac{d}{2}$ each and dielectric constant $K_{1}$and$K_{2}$. $d$ Is the plate separation of the capacitor. The capacity of the capacitor is
A) $\frac{2\varepsilon_{0}d}{A}\left( \frac{K_{1} + K_{2}}{K_{1}\ K_{2}}\ \right)$
B) $\frac{2\varepsilon_{0}A}{d}\left( \frac{K_{1}\ K_{2}}{K_{1} + \ K_{2}}\ \right)$
C) $\frac{2\varepsilon_{0}A}{A}\left( {K_{1} + \ K_{2}}\ \right)$
D) $\frac{2\varepsilon_{0}A}{d}\left( \frac{K_{1} + K_{2}}{K_{1}\ K_{2}}\ \right)$
Explanation
(b) $C_{1} = \frac{K_{1}\varepsilon_{0}A}{d/2}\ $=$\ \frac{2K_{1}\varepsilon_{0}A}{d}$
$C_{2}$=$\frac{2K_{2}\varepsilon_{0}A}{d}$
$\frac{1}{C_{s}} = \frac{1}{C_{1}\ } + \frac{1}{C_{2}} = \frac{d}{2K_{1}\varepsilon_{0}A} + \frac{d}{2K_{2}\varepsilon_{0}A}$
$= \frac{d}{2\varepsilon_{0}A}\left( \frac{K_{1} + K_{2}}{K_{1}K_{2}} \right)$
$C_{s} = \frac{2\varepsilon_{0}A}{d}\left( \frac{K_{1}K_{2}}{K_{1} + K_{2}} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.