A rectangular parallel-plate capacitor has a dielectric slab that partially fills the space between the plates as shown in fig. The energy of the capacitor with charge q is

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$\frac{q^{2} S}{2 ε_{0} b} [\frac{1}{Ka} - \frac{1}{(K - 1) x}]$

$\frac{q^{2} S}{2 ε_{0} b} [Ka - (K - 1) x]$

$\frac{q^{2} S}{2 ε_{0} b} \times \frac{1}{[Ka - (K - 1) x]}$

$\frac{q^{2} S}{ε_{0} b} \times \frac{1}{[Ka - (K - 1) x]}$

answer is A.

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Detailed Solution

$\begin{array}{l} C = \frac{ε_{0} bx}{S} + \frac{{Kε}_{0} b (a - x)}{S} \\ = \frac{ε_{0} b}{S} [x + K (a - x)] \\ = \frac{ε_{0} b}{S} [Ka - (K - 1) x] \\ ∴ U = \frac{1}{2} \frac{q^{2}}{C} = \frac{q^{2} S}{2 ε_{0} b} \frac{1}{[Ka - (K - 1) x]} \end{array}$