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Two identical capacitors $1$ and $2$ are connected in series. The capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0\ volts$ . The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charge stored in the capacitors before removing the slab and $Q'_1$ , and $Q'_2$ be the values after removing the slab. Then
$\frac{{Q{'_1}}}{{{Q_1}}} = \left( {\frac{{K + 1}}{K}} \right)$
$\frac{{Q{'_2}}}{{{Q_2}}} = \frac{{\left( {K + 1} \right)}}{2}$
$\frac{{Q{'_2}}}{{{Q_2}}} = \frac{{K + 1}}{{2K}}$
$\frac{{Q{'_1}}}{{{Q_1}}} = \frac{K}{2}$
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 $Q_{e q}^{\prime}=Q_{1}^{\prime}=Q_{2}^{\prime}$
$\therefore \frac{Q_{1}^{\prime}}{Q_{1}}=\frac{Q_{2}^{\prime}}{Q_{2}}=\frac{2 k}{k+1}$
