(A) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.Since they are connected with opposite polarities,the net charge on th

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(A) Initial charges on the capacitors are $Q_1 = CV$ and $Q_2 = (2C)(2V) = 4CV$.Since they are connected with opposite polarities,the net charge on the system is $Q_{net} = |Q_2 - Q_1| = |4CV - CV| = 3CV$.The equivalent capacitance of the system in parallel is $C_{eq} = C + 2C = 3C$.The common potential is $V_{com} = \frac{Q_{net}}{C_{eq}} = \frac{3CV}{3C} = V$.The final energy stored in the system is $U = \frac{1}{2} C_{eq} V_{com}^2 = \frac{1}{2} (3C) (V)^2 = \frac{3}{2} CV^2$.

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 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.

A
$3/2 \ CV^2$
B
$5/2 \ CV^2$
C
$7/3 \ CV^2$
D
$4/5 \ CV^2$

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Class 12 Questions

Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$ in $\mu C$?

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$A$ $400\, pF$ capacitor is charged with a $100\, V$ battery. After disconnecting the battery,this capacitor is connected to another uncharged $400\, pF$ capacitor. Find the energy loss.

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There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

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Two capacitors of capacitance $3 \ \mu F$ and $6 \ \mu F$ are charged to a potential of $12 \ V$. They are then connected by joining their positive plates together and their negative plates together. What will be the potential difference across each capacitor in $V$?

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$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ is

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In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$ in $\mu C$?

$A$ $400\, pF$ capacitor is charged with a $100\, V$ battery. After disconnecting the battery,this capacitor is connected to another uncharged $400\, pF$ capacitor. Find the energy loss.

There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

Two capacitors of capacitance $3 \ \mu F$ and $6 \ \mu F$ are charged to a potential of $12 \ V$. They are then connected by joining their positive plates together and their negative plates together. What will be the potential difference across each capacitor in $V$?

$A$ capacitor of capacity $C_1$ is charged to potential $V_1$ and then disconnected. An uncharged capacitor of capacity $C_2$ is connected in parallel with $C_1$. The resultant potential $V_2$ is

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