A slab of material of dielectric constant K has the same area as plate of parallel plate capacitor but has a thickness $(\frac{3}{4})$ d, where d is the separation of the plates then the capacity $(C_{o} = i n i t i a l c a p a c i t y)$

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$(\frac{k}{k + 3}) C_{o}$

$(\frac{3 k}{k + 4}) C_{o}$

$(\frac{2 k}{k + 3}) C_{o}$

$(\frac{4 k}{k + 3}) C_{o}$

answer is D.

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Detailed Solution

$\begin{array}{l} C = \frac{\in O A}{d_{-} t_{+} \frac{1}{k}} \\ = \frac{\in o A}{d - \frac{3 d}{4} + \frac{3 d}{4 k}} = (\frac{4 k}{k + 3}) C_{o} \end{array}$

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