(A) Initial energy of the system is $U_i = \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2$.When the capacitors are c

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(A) Initial energy of the system is $U_i = \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 = \frac{1}{2}CV^2 + 2CV^2 = \frac{5}{2}CV^2$.When the capacitors are connected in parallel,the common potential $V_c$ is given by $V_c = \frac{q_1 + q_2}{C_1 + C_2} = \frac{CV + 2CV}{C + C} = \frac{3CV}{2C} = \frac{3V}{2}$.The final energy of the system is $U_f = \frac{1}{2}(C + C)V_c^2 = C \left(\frac{3V}{2}\right)^2 = C \left(\frac{9V^2}{4}\right) = \frac{9}{4}CV^2$.The decrease in energy is $\Delta U = U_i - U_f = \frac{5}{2}CV^2 - \frac{9}{4}CV^2 = \frac{10}{4}CV^2 - \frac{9}{4}CV^2 = \frac{1}{4}CV^2$.

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

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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:

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A
$\frac{1}{4} CV^2$
B
$2 CV^2$
C
$\frac{1}{2} CV^2$
D
$\frac{3}{4} CV^2$

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Electrostatic Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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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:

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Energy $E$ is stored in a parallel plate capacitor $C_1$. An identical uncharged capacitor $C_2$ is connected to it,kept in contact with it for a while,and then disconnected. The energy stored in $C_2$ is

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