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In the arrangement shown in figure, dielectric constant $K_{1} = 2 a n d K_{2} = 3$ . If the capacitance across P and Q $C_{1} a n d C_{2}$ are respectively, then $C_{1} / C_{2}$ will be (the gaps shown are negligible)

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1 : 1

2 : 3

9 : 5

25 : 24

answer is D.

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Detailed Solution

$C_{1} = C_{R H S} + C_{L H S}$

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

$= \frac{ε_{0} A}{2 d} (K_{1} + K_{2}) = \frac{5 ε_{0} A}{2 d}$

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

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

$= \frac{12 ε_{0} A}{5 d}$

$∴ \frac{C_{1}}{C_{2}} = \frac{25}{24}$

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