Separation between the plates of a parallel plate capacitor is d and the area of each plate is A . When a slab of material of dielectric constant ‘k’ and thickness $t (t < d)$ is introduced between the plates its capacitance becomes

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$\frac{ε_{0} A}{d + t (1 - \frac{1}{k})}$

$\frac{ε_{0} A}{d + t (1 + \frac{1}{k})}$

$\frac{ε_{0} A}{d - t (1 - \frac{1}{k})}$

$\frac{ε_{0} A}{d - t (1 + \frac{1}{k})}$

answer is C.

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Detailed Solution

Potential difference between the plates $V = V_{1} + V_{2}$

$= \frac{σ}{ε_{0}} \times (d - t) + \frac{σ}{K ε_{0}} \times t (∵ E = \frac{σ}{ε_{0}} & v = E d)$

$\Rightarrow V = \frac{σ}{ε_{0}} (d - t + \frac{t}{k}) = \frac{Q}{A ε_{0}} (d - t + \frac{1}{K})$

Hence, capacitance $C = \frac{Q}{V} = \frac{Q}{\frac{Q}{A ε_{0}} (d - t + \frac{t}{K})}$

$= \frac{ε_{0} A}{(d - t + \frac{t}{K})} = \frac{ε_{0} A}{d - t (1 - \frac{1}{K})} .$

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