The parallel combination of two air-filled pa | vedclass

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(C) $1$. Initially,the capacitors are connected in parallel to a battery of voltage $V$. The total charge $Q_{total}$ stored in the system is given by

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$\frac{(n + 1)V}{K + n}$

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(C) $1$. Initially,the capacitors are connected in parallel to a battery of voltage $V$. The total charge $Q_{total}$ stored in the system is given by:$Q_{total} = C V + n C V = (n + 1) C V$$2$. After the battery is disconnected,the total charge $Q_{total}$ remains conserved in the system.$3$. When a dielectric of constant $K$ is inserted into the first capacitor,its new capacitance becomes $C' = K C$. The second capacitor remains unchanged with capacitance $n C$.$4$. Since the capacitors are in parallel,they share a common potential difference $V_c$. The total capacitance of the system becomes $C_{eq} = K C + n C = (K + n) C$.$5$. The new potential difference $V_c$ is given by:$V_c = \frac{Q_{total}}{C_{eq}} = \frac{(n + 1) C V}{(K + n) C} = \frac{(n + 1) V}{K + n}$

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