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Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in figure. Capacitor $2$ contains a dielectric slab of dielectric constant k as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q’_1$ and $Q’_2$. Then
$\frac{{{{Q'}_1}}}{{{Q_1}}} = \frac{{k + 1}}{k}$
$\frac{{{{Q'}_2}}}{{{Q_2}}} = \frac{{k + 1}}{2}$
$\frac{{{{Q'}_2}}}{{{Q_2}}} = \frac{{k + 1}}{{2k}}$
$\frac{{{{Q'}_1}}}{{{Q_1}}} = \frac{k}{2}$
Solution
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 $Q_{e q}^{\prime}=Q_{1}^{\prime}=Q_{2}^{\prime}$
$\therefore \frac{Q_{1}^{\prime}}{Q_{1}}=\frac{Q_{2}^{\prime}}{Q_{2}}=\frac{2 k}{k+1}$
