$A$ $10 \mu F$ capacitor is charged by a $100 V$ battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance $30 \mu F$. During this process,the electrostatic energy lost by the first capacitor is

$A$ capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2C$. The final charges on the capacitors are

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

$A$ 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.

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

Two hollow metallic spheres of radii $20 \ cm$ and $10 \ cm$ each carry a charge of $150 \ \mu C$. If they are connected by a conducting wire,the common potential is ...... $V$.