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Q.

Between the plates of a parallel plate capacitor of capacity C, two parallel plates, of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to ${\frac{1}{5}}^{t h}$ of the distance between the plates of the original capacitor, then the capacity of the new capacitor is

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a

$\frac{5}{3} C$

b

$\frac{3}{5} C$

c

$\frac{10 C}{3}$

d

$\frac{3 C}{10}$

answer is A.

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Detailed Solution

$C' = \frac{\in_{0} A}{(d - \frac{2 d}{5} + \frac{2 d}{5 K})}$ , Here K is $\infty$ .

we get $C' = \frac{5}{3} C$

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Between the plates of a parallel plate capacitor of capacity C, two parallel plates, of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to 15th of the distance between the plates of the original capacitor, then the capacity of the new capacitor is