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(B) Let the initial capacitances be $C_1 = C$ and $C_2 = C$. When a dielectric of constant $K = 4$ is inserted into the second capacitor,its new capac

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$80 \ V, 20 \ V$

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(B) Let the initial capacitances be $C_1 = C$ and $C_2 = C$. When a dielectric of constant $K = 4$ is inserted into the second capacitor,its new capacitance becomes $C_2' = K \cdot C = 4C$.In a series combination,the charge $Q$ on each capacitor is the same. Since $Q = CV$,we have $V = Q/C$,which implies $V \propto 1/C$.The ratio of capacitances is $C_1 : C_2' = C : 4C = 1 : 4$.Therefore,the ratio of potential differences is $V_1 : V_2 = 1/C_1 : 1/C_2' = 1/C : 1/4C = 4 : 1$.The total potential difference is $V = 100 \ V$.Potential difference across the first capacitor $(V_1)$: $V_1 = \frac{4}{4+1} 	imes 100 \ V = \frac{4}{5} 	imes 100 \ V = 80 \ V$.Potential difference across the second capacitor $(V_2)$: $V_2 = \frac{1}{4+1} 	imes 100 \ V = \frac{1}{5} 	imes 100 \ V = 20 \ V$.Thus,the potential differences are $80 \ V$ and $20 \ V$ respectively.

Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?

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