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(D) The given circuit is a balanced Wheatstone bridge for capacitors. Since all capacitors have equal capacitance $(C = 10\,\mu F)$,the potential diff

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(D) The given circuit is a balanced Wheatstone bridge for capacitors. Since all capacitors have equal capacitance $(C = 10\,\mu F)$,the potential difference across the central capacitor is zero. Therefore,no charge flows through the central capacitor,and it can be removed from the circuit. Now,the circuit consists of two parallel branches,each containing two capacitors in series. For the upper branch,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \implies C_1 = 5\,\mu F$. Similarly,for the lower branch,the equivalent capacitance $C_2$ is $5\,\mu F$. Since these two branches are in parallel,the total equivalent capacitance $C_{eq} = C_1 + C_2 = 5\,\mu F + 5\,\mu F = 10\,\mu F$.

Five capacitors of $10\,\mu F$ capacity each are connected to a $d.c.$ potential of $100\;V$ as shown in the figure. The equivalent capacitance between the points $A$ and $B$ will be equal to..........$\mu F$.

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