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Correct option is A)

As they are in series so the charge on each capacitors is equal to equivalent charge.i.eΒ

$Q_{eq}=q_{0}=C_{0}V_{0}$

With dielectric in (I), the capacitance of it becomes $kC_{0}$

now $C_{eq}=kC_{0}+C_{0}kC_{0}C_{0}β=k+1kβC_{0}$ and

$Q_{eq}=k+1kβC_{0}(2V_{0})=k+12kβq_{0}$

Thus the on each capacitor is $q_{β²}=Q_{eq}=k+12kβq_{0}=1+1/k2q_{0}β$

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