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(C) Let the five capacitors be connected in a pentagonal loop with vertices $P, Q, R, S, T$.Case $1$: Equivalent capacitance between $P$ and $R$ $(C_{

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$2:3$

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(C) Let the five capacitors be connected in a pentagonal loop with vertices $P, Q, R, S, T$.Case $1$: Equivalent capacitance between $P$ and $R$ $(C_{PR})$.When we look from $P$ to $R$,the path $P-Q-R$ consists of two capacitors in series,so their equivalent is $C/2$. The path $P-T-S-R$ consists of three capacitors in series,so their equivalent is $C/3$.These two branches are in parallel,so $C_{PR} = C/2 + C/3 = 5C/6$.Case $2$: Equivalent capacitance between $P$ and $Q$ $(C_{PQ})$.When we look from $P$ to $Q$,one path is the single capacitor $PQ$ with capacitance $C$. The other path is $P-T-S-R-Q$,which consists of four capacitors in series,so their equivalent is $C/4$.These two branches are in parallel,so $C_{PQ} = C + C/4 = 5C/4$.Ratio $= C_{PR} / C_{PQ} = (5C/6) / (5C/4) = (5/6) 	imes (4/5) = 4/6 = 2:3$.

Five capacitors,each of capacitance $C$,are connected as shown in the figure. The ratio of the equivalent capacitance between $P$ and $R$ to the equivalent capacitance between $P$ and $Q$ is:

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