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. Q₁ and Q₂ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are Q'₁ and Q'₂. Then

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$\frac{Q_{1}}{Q_{1}} = \frac{K + 1}{K}$

$\frac{Q_{2}}{Q_{2}} = \frac{K + 1}{2}$

$\frac{Q_{2}}{Q_{2}} = \frac{K + 1}{2 K}$

$\frac{Q_{2}}{Q_{2}} = \frac{K}{2}$

answer is C.

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Detailed Solution

$Q_{1}^{'} = Q_{2}^{'} = \frac{CE}{2}; Before the slab is removed$

$\begin{array}{l} C_{1} = C and C_{2} = kC; C_{net} = (\frac{k}{k + 1}) C \\ \frac{Q_{2}}{Q_{2}} = \frac{k + 1}{2 k} \end{array}$

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