$A$ parallel plate capacitor of capacitance $500 \ pF$ is charged with a $100 \ V$ supply. It is then disconnected from the supply and connected to another uncharged $500 \ pF$ capacitor. The electrostatic energy lost in this process is (in $\mu J$)

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 the positive ends are also connected,the decrease in energy of the combined system is

$A$ capacitor of $20\,\mu F$ is charged to $500\,V$ and connected in parallel with another capacitor of $10\,\mu F$ which is charged to $200\,V$. The common potential is.......$V$.

$A$ $10\ \mu F$ capacitor is charged to a potential difference of $1000\ V$. The terminals of the charged capacitor are disconnected from the power supply and connected to the terminals of an uncharged $6\ \mu F$ capacitor. What is the final potential difference across each capacitor in $V$?

Two capacitors of $10 \ pF$ and $20 \ pF$ are connected to $200 \ V$ and $100 \ V$ sources respectively. If they are connected by a wire,what is the common potential of the capacitors (in $V$)?

$A$ $2 \mu F$ 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 $\mu J$)