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If the area of each plate is $A$ and the successive separation are $3 d, 2 d$ and $d$ . Then find the equivalent capacitance across $A$ and $B$ .

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$- \frac{ε_{0} A}{d}$

$\frac{ε_{0} A}{4 d}$

None of the above

$\frac{4 ε_{0} A}{2}$

answer is A.

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Detailed Solution

In the figure, $C_{1} = \frac{ε_{0} A}{3 d}$

$C_{2} = \frac{ε_{0} A}{2 d}$

and $C_{3} = \frac{ε_{0} A}{d}$

Now, $C_{2}$ is short circuited so that $C_{1}$ and $C_{2}$ are in series.

$\Rightarrow C_{A B} = \frac{C_{1} C_{3}}{C_{1} + C_{3}} \Rightarrow C_{A B} = \frac{ε_{0} A}{4 d}$

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