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(D) In the steady state condition, capacitors act as open circuits, meaning no current flows through them. The circuit effectively becomes a series of

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$1 : n$

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(D) In the steady state condition, capacitors act as open circuits, meaning no current flows through them. The circuit effectively becomes a series of parallel branches connected to the voltage source $E$.For the first branch, the capacitor $C$ is connected directly across the voltage source $E$. Therefore, the potential difference across the first capacitor is $V_1 = E$. The charge stored in the first capacitor is $Q_1 = C \cdot V_1 = CE$.For the $n^{th}$ branch, the capacitor $nC$ is also connected directly across the voltage source $E$ because all branches are in parallel with the source. Therefore, the potential difference across the $n^{th}$ capacitor is $V_n = E$. The charge stored in the $n^{th}$ capacitor is $Q_n = (nC) \cdot V_n = nCE$.The ratio of the charge stored in the first capacitor to the charge stored in the last capacitor is $\frac{Q_1}{Q_n} = \frac{CE}{nCE} = \frac{1}{n}$.

For the shown situation, in the steady state condition, what is the ratio of the charge stored in the first capacitor to the charge stored in the last $(n^{th})$ capacitor?

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