(B) Initial energy stored in the two capacitors is:$U_i = \frac{1}{2} C V^2 + \frac{1}{2} C V^2 = C V^2$When the capacitors are connected with opposit

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(B) Initial energy stored in the two capacitors is:$U_i = \frac{1}{2} C V^2 + \frac{1}{2} C V^2 = C V^2$When the capacitors are connected with opposite polarities,the net charge on the system becomes:$Q_{net} = (+CV) + (-CV) = 0$Since the total charge is zero,the final potential difference across the capacitors is zero,and thus the final energy stored is:$U_f = 0$The loss of energy is given by:$\Delta U = U_i - U_f = C V^2 - 0 = C V^2$

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

Easy

Two identical capacitors are connected in parallel across a potential difference $V$. After they are fully charged,the positive plate of the first capacitor is connected to the negative plate of the second,and the negative plate of the first is connected to the positive plate of the other. The loss of energy will be

A
$\frac{1}{2} C V^2$
B
$C V^2$
C
$\frac{1}{4} C V^2$
D
$0$

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Class 12

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Electrostatic Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

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$A$ parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Another capacitor of capacitance $2C$ is charged to a potential $2V$ by another battery. After removing the batteries,they are connected in parallel such that the positive plate of one is connected to the negative plate of the other. Calculate the final energy of the system.

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$A$ capacitor of capacitance $C_1$ is charged to a potential $V_1$ and then disconnected. An uncharged capacitor of capacitance $C_2$ is connected in parallel with $C_1$. The resulting potential $V_2$ is:

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$A$ capacitor having capacitance $C$ is charged to a voltage $V$. It is then removed and connected in parallel with another identical capacitor which is uncharged. The new charge on each capacitor is now

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$A$ capacitor of capacitance $C$ is charged to a potential $V$. It is then disconnected and connected in parallel with an identical uncharged capacitor. Find the charge on each capacitor.

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Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

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