A parallel plate condenser initially has air as a medium between the plates. If a slab of dielectric constant 5 having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity will be ___

33.3 %

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66.7 %

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50 %

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75 %

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Solution

The correct option is B $66.7 %$
$C_{1} = \frac{ϵ_{0} A}{d}; C_{2} = \frac{ϵ_{0} A}{d - t + \frac{t}{k}} = \frac{ϵ_{0} A}{\frac{d}{2} + \frac{d}{10}}$
$C_{2} = \frac{10}{6} \frac{ϵ_{0} A}{d} = \frac{5 ϵ_{0} A}{3 d}$
$\frac{C_{2}}{C_{1}} = \frac{5}{3}; (\frac{C_{2}}{C_{1}} - 1) 100 % = (\frac{5}{3} - 1) 100 %$
$= 2 \times 33.3 %$
$= 66.67 %$

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A parallel plate condenser initially has air as a medium between the plates. If a slab of dielectric constant 5 having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity will be ___

The correct option is B 66.7%C1=ϵ0Ad;C2=ϵ0Ad−t+tk=ϵ0Ad2+d10 C2=106ϵ0Ad=5ϵ0A3d C2C1=53;(C2C1−1)100%=(53−1)100% =2×33.3% =66.67%