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(C) The circuit is symmetric. Due to symmetry,the potential at the nodes on the upper branch and the corresponding nodes on the lower branch are equal

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$C$

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(C) The circuit is symmetric. Due to symmetry,the potential at the nodes on the upper branch and the corresponding nodes on the lower branch are equal. Therefore,no current flows through the vertical capacitors,and they can be removed from the circuit.After removing the vertical capacitors,the circuit simplifies into two parallel branches. Each branch consists of capacitors in series: $C, 2C, 4C, 8C, \dots$The equivalent capacitance of one branch $(C_{branch})$ is given by:$\frac{1}{C_{branch}} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{4C} + \frac{1}{8C} + \dots$$\frac{1}{C_{branch}} = \frac{1}{C} \left( 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots ight)$This is a geometric series with first term $a = 1$ and common ratio $r = \frac{1}{2}$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.So,$\frac{1}{C_{branch}} = \frac{1}{C} 	imes 2 = \frac{2}{C} \Rightarrow C_{branch} = \frac{C}{2}$.Since there are two such branches in parallel,the total equivalent capacitance is:$C_{eq} = C_{branch} + C_{branch} = \frac{C}{2} + \frac{C}{2} = C$.

Find the equivalent capacitance of the circuit between points $A$ and $B$.

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