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(B) Initially,capacitors $B$ and $C$ are in parallel,and this combination is in series with capacitor $A$. Let the capacitance of each be $C_0$. The e

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Positive work is done on the given system by the external force

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(B) Initially,capacitors $B$ and $C$ are in parallel,and this combination is in series with capacitor $A$. Let the capacitance of each be $C_0$. The equivalent capacitance is $C_{eq} = \frac{C_0 \cdot (C_0 + C_0)}{C_0 + (C_0 + C_0)} = \frac{2C_0}{3}$.When a dielectric is inserted in $A$,its capacitance $C_A$ increases. Let $C_A = K C_0$ $(K > 1)$.The new equivalent capacitance is $C'_{eq} = \frac{C_A \cdot 2C_0}{C_A + 2C_0}$. Since $C_A > C_0$,$C'_{eq} > \frac{2C_0}{3}$.The total charge $Q_{total} = C'_{eq} \varepsilon$ increases. Since $Q_{total}$ is the charge on $A$,the charge on $A$ increases.The potential difference across $A$ is $V_A = \varepsilon - V_{BC}$,where $V_{BC}$ is the potential across the parallel combination of $B$ and $C$. As $C_A$ increases,$V_A$ decreases,so $V_{BC}$ increases.Since $V_{BC}$ increases,the charge on $B$ $(q_B = C_0 V_{BC})$ and $C$ $(q_C = C_0 V_{BC})$ both increase.Since the battery supplies more charge,the chemical energy of the battery changes.The force between the plates of a capacitor is $F = \frac{q^2}{2 \epsilon_0 A}$. Since $q_B$ and $q_C$ increase,the force between the plates of $B$ and $C$ increases.Work done by the external force $W_{ext} = \Delta U + W_{batt}$. Since the energy stored increases and the battery does positive work,the external force does positive work to insert the dielectric.

Three capacitors $A, B$ and $C$ are connected with a battery of $emf$ $\varepsilon$. All capacitors are identical initially. If a dielectric slab is inserted between the plates of capacitor $A$ slowly with the help of an external force,then:

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