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(C) The arrangement consists of three parallel plate capacitors connected in parallel.The capacitance of a parallel plate capacitor is given by $C = \

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$23$

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(C) The arrangement consists of three parallel plate capacitors connected in parallel.The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.From the figure, the separation distances between the plates are $b$, $3b$, and $5b$.Therefore, the individual capacitances are:$C_1 = \frac{\varepsilon_0 A}{b}$$C_2 = \frac{\varepsilon_0 A}{3b}$$C_3 = \frac{\varepsilon_0 A}{5b}$Since they are in parallel, the equivalent capacitance is:$C_{eq} = C_1 + C_2 + C_3$$C_{eq} = \frac{\varepsilon_0 A}{b} + \frac{\varepsilon_0 A}{3b} + \frac{\varepsilon_0 A}{5b}$$C_{eq} = \frac{\varepsilon_0 A}{b} \left( 1 + \frac{1}{3} + \frac{1}{5} \right)$$C_{eq} = \frac{\varepsilon_0 A}{b} \left( \frac{15 + 5 + 3}{15} \right) = \frac{23}{15} \frac{\varepsilon_0 A}{b}$Comparing this with $\frac{ x }{15} \frac{\varepsilon_{0} A }{ b }$, we get $x = 23$.

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