$A$ $400\, pF$ capacitor is charged with a $100\, V$ battery. After disconnecting the battery,this capacitor is connected to another uncharged $400\, pF$ capacitor. Find the energy loss.

Two identical capacitors each of capacitance $5 \mu F$ are charged to potentials $2 kV$ and $1 kV$ respectively. Their negative ends are connected together. When the positive ends are also connected together,the loss of energy of the system is

$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.

$A$ $10\,\mu F$ capacitor is fully charged to a potential difference of $50\, V$. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now, the potential difference across them becomes $20\, V$. The capacitance of the second capacitor is $\dots \mu F$.

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

Two capacitors of capacitances $3 \mu F$ and $6 \mu F$ are charged to a potential of $12 V$ each. They are now connected to each other,with the positive plate of each joined to the negative plate of the other. The potential difference across each will be $volt$.