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The capacitor shown has square plates of side L and has gap d between the plates. If the dielectric constant of the material of the slab is k, the capacitance of the capacitor is

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$C_{eq} = \frac{ε_{0} L^{2} (1 + 2 K)}{d (1 + 2 K)}$

$C_{eq} = \frac{ε_{0} L^{2} (1 + 2 K)}{d (1 + K)}$

$C_{eq} = \frac{ε_{0} L^{2} (1 + 3 K)}{d (K + 1)}$

$C_{eq} = \frac{ε_{0} L^{2} (1 + 3 K)}{d (1 + 2 K)}$

answer is B.

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Detailed Solution

On the side with the dielectric slab :

$\frac{1}{C} = \frac{d}{ε_{0} KA} + \frac{2 d}{ε_{0} A}$

The total equivalent capacitance is

$\Rightarrow C_{eq} = \frac{ε_{0} A}{d} + \frac{ε_{0} KA}{d (1 + 2 K)}$

$C_{eq} = \frac{ε_{0} L^{2} (1 + 3 K)}{d (1 + 2 K)}$

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