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Increasing Capacitance Even More

In the given ...

Question

In the given arrangement for dielectric constants $K_{1} = 2$ and $K_{2} = 3$ , the capacitance across $P$ and $Q$ are $C_{1}$ and $C_{2}$ respectively in figure $(a)$ and $(b)$ . Then $C_{1} / C_{2}$ will be

1 : 1

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2 : 3

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24 : 25

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25 : 24

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Solution

The correct option is D $25 : 24$
From figure $(a)$ , capacitors are in parallel.
So,

$C_{1} = \frac{K_{1} ϵ_{0} (A / 2)}{d} + \frac{K_{2} ϵ_{0} (A / 2)}{d}$

$\Rightarrow C_{1} = \frac{ϵ_{0} A}{2 d} (K_{1} + K_{2}) . . . . . . . . . (1)$

From figure $(b)$ , capacitors are in series.
So,

$\frac{1}{C_{2}} = \frac{1}{\frac{K_{1} ϵ_{0} A}{(d / 2)}} + \frac{1}{\frac{K_{2} ϵ_{0} A}{(d / 2)}}$

$\Rightarrow \frac{1}{C_{2}} = \frac{1}{\frac{2 K_{1} ϵ_{0} A}{d}} + \frac{1}{\frac{2 K_{2} ϵ_{0} A}{d}}$

$\Rightarrow C_{2} = \frac{2 ϵ_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}}) . . . . . . . (2)$

From eq. $(1) & (2)$

$\frac{C_{1}}{C_{2}} = \frac{\frac{ϵ_{0} A (K_{1} + K_{2})}{2 d}}{\frac{2 ϵ_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})}$

$\Rightarrow \frac{C_{1}}{C_{2}} = \frac{{(K_{1} + K_{2})}_{1}^{2}}{4 K_{1} K_{2}} = \frac{{(2 + 3)}^{2}}{4 \times 2 \times 3}$

$\Rightarrow \frac{C_{1}}{C_{2}} = \frac{25}{24} = 25 : 24$

Hence, option $(d)$ is correct.

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