Q98 / 160 · JIPMER 2013 · Physics
A parallel plate condenser is filled with two dielectrics as shown in figure. Area of each pate is $A \mathrm{~m}^2$ and the separation is $d$ metre. The dielectric constants are $K_1$ and $K_2$ respectively. Its capacitance in farad will be


- A$\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1+K_2}{K_1 K_2}\right)$
- Correct answerB$\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)$
- C$\frac{\varepsilon_0 A}{d}\left(\frac{K_1+K_2}{2 K_1 K_2}\right)$
- D$\frac{\varepsilon_0 A K_1 K_2}{2\left(d_2 K_1+d_1 K_2\right)}$
Solution
From the formula
$\begin{aligned} C & =\frac{\varepsilon_0 A}{\frac{1}{k_1}+\frac{1}{k_2}}=\frac{\varepsilon_0 A}{t}\left(\frac{k_1 k_2}{k_1+k_2}\right) \\ & =\frac{2 \varepsilon_0 A}{d}\left(\frac{k_1 k_2}{k_1+k_2}\right)\end{aligned}$
Hence, the correct answer is option: 'B'.