A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in fig. The separation of the plate is d. If b = d/2.
then the ratio of capacities of the capacitor after and before inserting the slab will be

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$1 : \sqrt{2}$

$2 : 1$

$\sqrt{2} : 1$

1 : 1

answer is B.

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Detailed Solution

Capacity of condenser before insertion of plate $C = \frac{ε_{0} A}{d}$
Capacity of condenser after insertion of plate $C^{'} = \frac{ε_{0} A}{(d / 2)} = \frac{2 ε_{0} A}{d} ∴ \frac{C^{'}}{C} = \frac{2}{1}$

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