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(B) Initially,both capacitors have capacitance $C$. After inserting the dielectric slab of constant $K=4$ in the second capacitor,its new capacitance

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$80\,V, 20\,V$

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(B) Initially,both capacitors have capacitance $C$. After inserting the dielectric slab of constant $K=4$ in the second capacitor,its new capacitance becomes $C_2 = KC = 4C$. The first capacitor remains $C_1 = C$.Since they are in series,the equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{C 	imes 4C}{C + 4C} = \frac{4C}{5}$.The total charge stored in the series combination is $Q = C_{eq} V = \frac{4C}{5} 	imes 100 = 80C$.In a series circuit,the charge $Q$ is the same for both capacitors.The potential difference across the first capacitor is $V_1 = \frac{Q}{C_1} = \frac{80C}{C} = 80\,V$.The potential difference across the second capacitor is $V_2 = \frac{Q}{C_2} = \frac{80C}{4C} = 20\,V$.Thus,the potential differences are $80\,V$ and $20\,V$ respectively.

Two identical parallel plate capacitors are connected in series to a battery of $100\,V$. $A$ dielectric slab of dielectric constant $4.0$ is inserted between the plates of the second capacitor. The potential difference across the capacitors will now be respectively:

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