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Arrange the following in decreasing order of effective capacitance between P and Q

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II, IV, III, I

IV, III, II, I

I, II, III, IV

I, III, II, IV

answer is B.

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Detailed Solution

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

$C_{eq} = \frac{5 ε_{0} A}{2 d}$

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

$C_{1} = \frac{4 ε_{0} A}{d} & C_{2} = \frac{6 ε_{0} A}{d}$

$C_{eq} = \frac{12 ε_{0} A}{5 d}$

3) $C_{1} = \frac{K_{1} ε_{0} A / 2}{d / 2}; C_{2} = \frac{K_{2} ε_{0} A / 2}{d / 2}; C_{3} = \frac{K_{3} ε_{0} A}{2 d}$

$C_{1} = \frac{2 ε_{0} A}{d}; C_{2} = \frac{3 ε_{0} A}{d}; C_{3} = \frac{ε_{0} A}{d}$

$C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} + C_{3} = \frac{11 ε_{0} A}{5 d}$

4) $C_{eq} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} & C_{1} = \frac{K_{1} ε_{0} A}{d / 2}; C_{2} = \frac{ε_{0} A}{d / 2}$

$C_{1} = \frac{4 ε_{0} A}{d}; C_{2} = \frac{2 ε_{0} A}{d}$

$C_{eq} = \frac{4 ε_{0} A}{3 d}$

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