Question

A parallel plate condenser with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the given below. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. The capacitance will be:-

• A. $\frac{\left(\epsilon \right)_{0} A}{ d}\left(K_{1} + K_{2}\right)$
• B. $\frac{\left(\epsilon \right)_{0} A}{ d}\left(\frac{K_{1} + K_{2}}{K_{1} K_{2}}\right)$
• C. $\frac{2 \left(\epsilon \right)_{0} A}{ d}\left(\frac{K_{1} K_{2}}{ K_{1} + K_{2}}\right)$
• D. $\frac{2 \left(\epsilon \right)_{0} A}{d}\left(\frac{K_{1} + K_{2}}{ K_{1} K_{2}}\right)$

Answer 1

Answer: (C)

The combination is equivalent to two capacitors in series, each with plate area $A$ and separation $d/2$
$C_{1}=\frac{K_{1} \left(\epsilon \right)_{0} A}{ \left(d / 2\right)}=\frac{2 K_{1} \left(\epsilon \right)_{0} A}{ d}$
$C_{2}=\frac{K_{2} \left(\epsilon \right)_{0} A}{\left(d / 2\right)}=\frac{2 K_{2} \left(\epsilon \right)_{0} A}{ d}$
Further, $\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{2 \left(\epsilon \right)_{0} A}\left(\frac{1}{ K_{1}} + \frac{1}{ K_{2}}\right)$