Question

A parallel plate capacitor has the space between its plates filled by the two slabs of thickness $\frac{d}{2}$ each and dielectric constants $K_{1}$ 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. $\frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right)$

Answer 1

Answer: (B)

$C_{1}=\frac{K_{1} \varepsilon_{0} A}{d / 2}=\frac{2 K_{1} \varepsilon_{0} A}{d}$ and $C_{2}=\frac{2 K_{2} \varepsilon_{0} A}{d}$
$\therefore \frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{2 K_{1} \varepsilon_{0} A}+\frac{d}{2 K_{2} \varepsilon_{0} A}$
$=\frac{d}{2 \varepsilon_{0} A}\left(\frac{K_{1}+K_{2}}{K_{1} K_{2}}\right)$
$\Rightarrow$ Capacity, $C_{s}=\frac{2 \varepsilon_{0} A}{d}\left(\frac{K_{1} K_{2}}{K_{1}+K_{2}}\right)$