Two identical capacitors are connected in series. Charge on each capacitor is q_o. A dielectric slab is now introduced between the plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge in each capacitor will now be

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$\frac{q_{0}}{1 + k}$

$\frac{2 q_{0}}{1 + k}$

$\frac{q_{0}}{1 + \frac{1}{k}}$

$\frac{2 q_{0}}{1 + \frac{1}{k}}$

answer is A.

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Detailed Solution

$\begin{array}{l} q_{0} = \frac{C_{0} V_{0}}{2} and q = C_{s} V \\ \Rightarrow q = [(\frac{k}{k + 1}) C_{0}] (V_{0}) \\ q = \frac{C_{0} V_{0}}{\frac{K + 1}{K}} = \frac{2 q_{0}}{1 + \frac{1}{K}} q = \frac{C_{0} V_{0}}{\frac{K + 1}{K}} = \frac{2 q_{0}}{1 + \frac{1}{K}} \end{array}$