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(A) Let the equivalent capacitance of the ladder between points $A$ and $B$ be $C_{eq}$.If we add one more section to the ladder,the equivalent capaci

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$4$

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(A) Let the equivalent capacitance of the ladder between points $A$ and $B$ be $C_{eq}$.If we add one more section to the ladder,the equivalent capacitance remains unchanged because the ladder is infinite or effectively behaves as such for the condition to hold.Looking at the circuit,the first section consists of a $2\,\mu F$ capacitor in parallel with the rest of the ladder,which is in series with a $4\,\mu F$ capacitor.Thus,$C_{eq} = 2 + \frac{4 	imes C_{eq}}{4 + C_{eq}}$.Solving for $C_{eq}$:$C_{eq} - 2 = \frac{4 C_{eq}}{4 + C_{eq}}$$(C_{eq} - 2)(4 + C_{eq}) = 4 C_{eq}$$4 C_{eq} + C_{eq}^2 - 8 - 2 C_{eq} = 4 C_{eq}$$C_{eq}^2 - 2 C_{eq} - 8 = 0$$(C_{eq} - 4)(C_{eq} + 2) = 0$Since capacitance cannot be negative,$C_{eq} = 4\,\mu F$.For the ladder to be independent of the number of sections,the terminating capacitor $C$ must be equal to this equivalent capacitance $C_{eq}$.Therefore,$C = 4\,\mu F$.

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