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(D) In the steady state condition,the capacitors act as open circuits,meaning no current flows through them. Since all branches are connected in paral

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$1 : n$

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(D) In the steady state condition,the capacitors act as open circuits,meaning no current flows through them. Since all branches are connected in parallel across the battery of $EMF$ $E$,the potential difference across each capacitor is equal to the $EMF$ of the battery,$E$. The charge stored in the first capacitor $(C_1 = C)$ is $Q_1 = C 	imes E$. The charge stored in the $n^{th}$ capacitor $(C_n = nC)$ is $Q_n = (nC) 	imes E$. Therefore,the ratio of the charge stored in the first capacitor to the charge stored in the $n^{th}$ capacitor is $\frac{Q_1}{Q_n} = \frac{CE}{nCE} = \frac{1}{n}$.

For the shown situation,in the steady state condition,the ratio of the charge stored in the first capacitor to the charge stored in the $n^{th}$ capacitor is:

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