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 Figure

  • A
    $\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1+K_2}{K_1 K_2}\right)$
  • Correct answer
    B
    $\frac{2 \varepsilon_0 A}{d}\left(\frac{K_1 K_2}{K_1+K_2}\right)$
  • 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'.

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