$A$ $4 \mu F$ capacitor is charged by a $200 \ V$ battery. It is then disconnected from the supply and connected to another uncharged $2 \mu F$ capacitor. During the process,the loss of energy (in $J$) is:

$A$ 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

The capacities of two conductors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2$. If they are connected by a thin wire,then the loss of energy will be given by

$A$ body of capacity $4\,\mu F$ is charged to $80\,V$ and another body of capacity $6\,\mu F$ is charged to $30\,V$. When they are connected,the energy lost by the $4\,\mu F$ capacitor is.......$mJ$.

$A$ 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:

Two conducting spheres of radii $5\, cm$ and $10\, cm$ are given a charge of $15\,\mu C$ each. After the two spheres are joined by a conducting wire, the charge on the smaller sphere is.......$\mu C$.