$A$ capacitor is charged with a battery and the energy stored is $U$. After disconnecting the battery,another identical uncharged capacitor is connected in parallel with it. The total energy of the system of capacitors is . . . . . . .

$C_1$ धारिता वाले एक संधारित्र को $V_1$ विभव तक आवेशित किया जाता है और फिर वियोजित कर दिया जाता है। $C_2$ धारिता वाले एक अनावेशित संधारित्र को $C_1$ के साथ समांतर क्रम में जोड़ा जाता है। परिणामी विभव $V_2$ है:

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 $\mu C$?

$A$ 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 $CV^2$)

Consider two uncharged capacitors of equal capacitance $200 \text{ pF}$. One of them is charged by a $100 \text{ V}$ supply and disconnected. Now, this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

$A$ conducting sphere $S_1$ of radius $r_1 = 3 \text{ cm}$ is connected by a conducting wire to another conducting sphere $S_2$ of radius $r_2 = 2 \text{ cm}$. Before they are connected,$S_1$ carries a charge of $10 \text{ units}$. The electric potential at a point which is at a distance $4 \text{ cm}$ from the centre of $S_1$ and a distance $3 \text{ cm}$ from the centre of $S_2$ is: