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Question

The capacitor $1$ has capacitance $C$ and as dielectric contains in capacitor $2$ so its capacitance becomes $k C$

The net capacitance $C_{e q}=\fr

Vedclass · Accepted Answer

$\frac{{{{Q'}_2}}}{{{Q_2}}} = \frac{{k + 1}}{{2k}}$

Vedclass · Answer

The capacitor $1$ has capacitance $C$ and as dielectric contains in capacitor $2$ so its capacitance becomes $k C$

The net capacitance $C_{e q}=\frac{C \cdot k C}{C+k C}=\frac{C k}{1+k}$

and $Q_{e q}=C_{e q} E$

When dielectric is reemoved from capacitor $2,$ its capacitance becomes $C$.

now net capacitance $C_{e q}^{\prime}=\frac{C . C}{C+C}=\frac{C}{2}$

and $Q_{e q}^{\prime}=C_{e q}^{\prime} E=\frac{C E}{2}$

When the capacitors are connected in series, they each have the same charge as the

net capacitance. Thus, $Q_{e q}=Q_{1}=Q_{2}$ and $Q_{e q}^{\prime}=Q_{1}^{\prime}=Q_{2}^{\prime}$

$	herefore \frac{Q_{1}^{\prime}}{Q_{1}}=\frac{Q_{2}^{\prime}}{Q_{2}}=\frac{2 k}{k+1}$

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