In question , the capacitor is connected to a battery with terminal voltage V This connection is maintained while the dielectric slab is held in place by the application of a force F. The required force is plates of one of the capacitors so as to fill the gap, the battery remaining connected. The charge on each capacitor will be

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$\frac{2 q_{0}}{[1 + (1 / K)]}$

$\frac{q_{0}}{1 + (1 / K)}$

$\frac{2 q_{0}}{(1 + K)}$

$\frac{q_{0}}{(1 + K)}$

answer is A.

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Detailed Solution

Before the introduction of dielectric, the charge on each capacitor will be the same because they are connected in series.
Initial charge on each capacitor q₀= C₀V₀ When dielectric is introduced, C₁ = C₀ and C₂ = K C₀
$C = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{C_{0} \times C_{0} K}{C_{0} + C_{0} K} = \frac{C_{0} K}{1 + K}$
Charge on each capacitor
$q = C (2 V_{0}) = (\frac{C_{0} K}{1 + K}) (2 V_{0}) = \frac{2 q_{0}}{[1 + (1 / K)]}$