The space between the plates of two parallel plate capacitors is filled by two dielectric slabs having dielectric constants $k_{1} and k_{2}$ as shown in fig-(1) and (2). Plate area of each capacitor is A and plate separation is d. If $c_{1} and c_{2}$ be the capacitances of the capacitors in arrangements (1) and (2) respectively, then $\frac{c_{1}}{c_{2}}$ is

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$\frac{k_{1} k_{2}}{(k_{1} + k_{2})}$

$\frac{2 k_{1} k_{2}}{{(k_{1} + k_{2})}^{2}}$

$\frac{{(k_{1} + k_{2})}^{2}}{4 k_{1} k_{2}}$

$\frac{(k_{1} + k_{2})}{2 k_{1} k_{2}}$

answer is D.

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Detailed Solution

$c_{1} = \frac{ε_{0} k_{1} (A / 2)}{d} + \frac{ε_{0} k_{2} (A / 2)}{d} = \frac{ε_{0} (k_{1} + k_{2}) A}{2 d}$

$\frac{1}{c_{2}} = \frac{d / 2}{k_{1} ε_{0} A} + \frac{d / 2}{k_{2} ε_{0} A} = \frac{d}{2 ε_{0} A} x (\frac{k_{1} + k_{2}}{k_{1} k_{2}})$

$\Rightarrow c_{2} = \frac{2 ε_{0} A}{d} (\frac{k_{1} k_{2}}{k_{1} + k_{2}})$

$∴ \frac{c_{1}}{c_{2}} = \frac{ε_{0} (k_{1} + k_{2}) A}{2 d} x \frac{d (k_{1} + k_{2})}{2 ε_{0} {Ak}_{1} k_{2}} = \frac{{(k_{1} + k_{2})}^{2}}{4 k_{1} k_{2}}$

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