$A$ 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

$A$ $4 \;\mu F$ capacitor is charged by a $200 \;V$ supply. It is then disconnected from the supply and connected to another uncharged $2 \;\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Two metallic spheres of radii $1 \, cm$ and $3 \, cm$ are given,with charges of $-1 \times 10^{-2} \, C$ and $5 \times 10^{-2} \, C$ respectively. If these are connected by a conducting wire,the final charge on the bigger sphere is

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging batteries are then removed,and the capacitors are connected in parallel such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

$A$ 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 $\text{ mJ}$)

$A$ 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.