A parallel plate capacitor has capacitance $C$ . If it is equally filled the parallel of materials of dielectric constant $K_{1}$ and $K_{2}$ its capacity becomes $C_{1}$ . The ratio of $C_{1}$ and $C$ is

K_{1} + K_{2}

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\frac{K_{1} K_{2}}{K_{1} + K_{2}}

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\frac{K_{1} + K_{2}}{K_{1} K_{2}}

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\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}

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Solution

The correct option is D $\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$
Capacitance, $C_{A} = \frac{K_{1} ε_{0} A}{\frac{d}{2}}$ , $C_{B} = \frac{K_{2} ε_{0} A}{\frac{d}{2}}$
$C_{e q} = \frac{C_{1}}{C_{2}} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$
$= \frac{C_{A} C_{B}}{C_{A} + C_{B}} = (\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}) \frac{ε_{0} A}{d} (∵ e = \frac{ε_{0} A}{d})$

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A parallel plate capacitor has capacitance C. If it is equally filled the parallel of materials of dielectric constant K1 and K2 its capacity becomes C1. The ratio of C1 and C is

The correct option is D 2K1K2K1+K2Capacitance, CA=K1ε0Ad2, CB=K2ε0Ad2Ceq=C1C2=2K1K2K1+K2=CACBCA+CB=(2K1K2K1+K2)ε0Ad(∵e=ε0Ad)

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