Question

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

• A. $\frac{K_1{K}_2}{K_1+K_2}$
• B. $K_1+K_2$
• C. $\frac{K_1+K_2}{K_1{K}_2}$
• D. $\frac{2K_1{K}_2}{K_1+K_2}$

Answer 1

Answer: (D)

Capacitance, $C_A=\frac{K_1{\epsilon }_{0}A}{\frac{d}{2}}$, $C_B=\frac{K_2{\epsilon }_{0}A}{\frac{d}{2}}$
$C_{eq}=\frac{C_1}{C_2}=\frac{2K_1{K}_2}{K_1+K_2}$
$=\frac{C_A{C}_B}{C_A+C_B}=\left(\frac{2K_1{K}_2}{K_1+K_2}\right)\frac{{\epsilon }_{0}A}{d}(\because e=\frac{{\epsilon }_{0}A}{d})$