(C) Initial energy of the system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2 = \frac{1}{2}C(V_1^2 + V_2^2)$.When connected with like plates togeth

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(C) Initial energy of the system is $U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2 = \frac{1}{2}C(V_1^2 + V_2^2)$.When connected with like plates together,the common potential is $V = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}$.The final energy of the system is $U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1 + V_2)^2$.The decrease in energy is $\Delta U = U_i - U_f = \frac{1}{2}C(V_1^2 + V_2^2) - \frac{C}{4}(V_1 + V_2)^2$.$\Delta U = \frac{C}{4} [2V_1^2 + 2V_2^2 - (V_1^2 + V_2^2 + 2V_1V_2)] = \frac{C}{4}(V_1^2 + V_2^2 - 2V_1V_2) = \frac{C}{4}(V_1 - V_2)^2$.

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

Medium

One of the two identical capacitors having the same capacitance $C$ is charged to a potential $V_1$ and the other is charged to a potential $V_2$. If they are connected with their like plates together,then the decrease in the electrostatic potential energy of the combined system is

AP EAMCET 2025 All AP EAMCET

A
$\frac{C}{4}(V_1^2 - V_2^2)$
B
$\frac{C}{4}(V_1^2 + V_2^2)$
C
$\frac{C}{4}(V_1 - V_2)^2$
D
$\frac{C}{4}(V_1 + V_2)^2$

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Electrostatic Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

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Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

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Two metal spheres have radii of $20\, cm$ and $10\, cm$ respectively, and each sphere carries a charge of $150\, \mu C$. After connecting them with a conducting wire, the common potential on them will be:

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$A$ parallel plate capacitor of capacitance $10 \mu F$ is charged by a $220 \text{ V}$ supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance $12 \mu F$. The loss of electrostatic energy in this process is (in $\text{ mJ}$)

MediumTS EAMCET 2025

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Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$,respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is $.....CV^2$.

DifficultJEE MAIN 2020

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One of the two identical capacitors having the same capacitance $C$ is charged to a potential $V_1$ and the other is charged to a potential $V_2$. If they are connected with their like plates together,then the decrease in the electrostatic potential energy of the combined system is

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$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

Two metal spheres have radii of $20\, cm$ and $10\, cm$ respectively, and each sphere carries a charge of $150\, \mu C$. After connecting them with a conducting wire, the common potential on them will be:

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