Q. A parallel plate capacitor has capacitance $C.$ If it is equally filled with parallel layers of materials of dielectric constants $K_{1}$ and $K_{2}$ its capacity becomes $C_{1}.$ The ratio of $C_{1}$ to $C$ is
A) $K_{1} + K_{2}$
B) $\frac{K_{1}K_{2}}{K_{1} - K_{2}}$
C) $\frac{K_{1} + K_{2}}{K_{1}K_{2}}$
D) $\frac{2K_{1}K_{2}}{K_{1} + K_{2}}$
Explanation
(d) $C_{A} = \frac{K_{1}\varepsilon_{0}A}{d/2},C_{B} = \frac{K_{2}\varepsilon_{0}A}{d/2}$
$\therefore C_{eq} = \frac{C_{A}C_{B}}{C_{A} + C_{B}} = \left( \frac{2K_{1}K_{2}}{K_{1} + K_{2}} \right)\frac{\varepsilon_{0}A}{d}$
$\left\lbrack \therefore c = \frac{\varepsilon_{0}A}{d} \right\rbrack\ \therefore\frac{C_{1}}{C} = \frac{2K_{1}K_{2}}{K_{1} + K_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.