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Q.

The two metallic plates of radius r are placed at a distance d apart and its capacity is C. If a plate of radius r/2 and thickness d of dielectric constant 6 is placed between the plates of the condenser, then its capacity will be

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a

3C/7

b

7C/3

c

9C/4

d

7C/2

answer is D.

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Detailed Solution

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Area of the given metallic plate $A = {πr}^{2}$

Area of the dielectric plate $A^{'} = π {(\frac{r}{2})}^{2} = \frac{A}{4}$

Uncovered area of the metallic plates

$\begin{array}{l} A " A - A^{'} = A - \frac{A}{4} = \frac{3 A}{4} \\ C_{eq} = C^{'} + C^{''} = \frac{{Kε}_{0} (A / 4)}{d} + \frac{ε_{0} (3 A / 4)}{d} \\ = \frac{ε_{0} A}{d} (\frac{K}{4} + \frac{3}{4}) = \frac{ε_{0} A}{d} (\frac{6}{4} + \frac{3}{4}) = \frac{9}{4} C \\ (∵ C = \frac{ε_{0} A}{d}) \end{array}$

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