Two capacitors of $1\,\mu F$ and $2\,\mu F$ are charged to $1200\,V$. If they are connected in parallel such that the positive plate of one is connected to the negative plate of the other,what will be the new charge on each capacitor?

$A$ 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.......$mJ$.

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When positive ends are also connected,the decrease in energy of the combined system is

$A$ $2 \mu F$ capacitor is charged to a potential $V_1 = 10 \ V$. Another $4 \mu F$ capacitor is charged to a potential $V_2 = 20 \ V$. The two capacitors are then connected in a single loop,with the positive plate of one connected to the negative plate of the other. What heat is evolved in the circuit? (in $\mu J$)

Two capacitors $C_1$ and $C_2$ are charged to $120 \ V$ and $200 \ V$ respectively. It is found that connecting them together the potential on each one can be made zero. Then

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