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The plates in a parallel plate capacitor are separated by a distance d with air as the medium between the plates . In order to increase the capacity by 66% a dielectric slab of dielectric constant 5 is introduced between the plates. What is the thickness of the dielectric slab.

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$\frac{d}{4}$

$\frac{5 d}{8}$

$d$

$\frac{d}{2}$

answer is B.

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Detailed Solution

$C = \frac{A ε_{0}}{d}$

$C^{1} = \frac{A ε_{0}}{d - t + \frac{t}{5}}$

$\frac{C}{C^{1}} = \frac{d - t + \frac{t}{5}}{d}$

$C^{1} = \frac{166}{100} C$

$\frac{100}{166} = \frac{d - \frac{4 t}{5}}{d}$

$t = \frac{d}{2}$

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