(C) Initial charge on the first capacitor: $q_1 = CV$. Initial charge on the second capacitor: $q_2 = (3C)(3V) = 9CV$. Since they are connected with o

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(C) Initial charge on the first capacitor: $q_1 = CV$. Initial charge on the second capacitor: $q_2 = (3C)(3V) = 9CV$. Since they are connected with opposite polarities, the net charge $Q_{net} = q_2 - q_1 = 9CV - CV = 8CV$. The equivalent capacitance of the parallel combination is $C_{eq} = C + 3C = 4C$. The common potential $V'$ is given by $V' = Q_{net} / C_{eq} = 8CV / 4C = 2V$. The final energy $U_f$ of the configuration is $U_f = (1/2) C_{eq} (V')^2$. Substituting the values: $U_f = (1/2) (4C) (2V)^2 = (1/2) (4C) (4V^2) = 8CV^2$.

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

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$A$ parallel plate capacitor of capacitance '$C$' is connected to a battery and charged to a potential '$V$'. Another capacitor of capacitance '$3C$' is charged to a potential '$3V$'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (in $CV^2$)

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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MediumJEE MAIN 2020

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MediumAIPMT 2002

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Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2V$. The negative ends of both are connected together. When the positive ends are also joined together,the decrease in energy of the combined system is:

DifficultJEE MAIN 2024

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