$A$ capacitor of capacitance $C$ and potential $V$ has energy $E$. It is connected to another capacitor of capacitance $2C$ and potential $2V$. Then the loss of energy is $\frac{x}{3} E$,where $x$ is . . . . . . .

$A$ 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 $\mu J$)

Consider two uncharged capacitors of equal capacitance $200 \text{ pF}$. One of them is charged by a $100 \text{ V}$ supply and disconnected. Now,this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

$A$ $10 \mu F$ 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 capacity $4 \, \mu F$ is charged to $80 \, V$ and another capacitor of capacity $6 \, \mu F$ is charged to $30 \, V$. When they are connected, the energy lost by the $4 \, \mu F$ capacitor is: (in $ \, mJ$)

$A$ $2\,\mu F$ capacitor has a potential difference of $200\;V$ across its terminals. It is disconnected from the battery and then another uncharged capacitor is connected in parallel to it. If the potential difference becomes $20\;V$,the capacitance of the second capacitor is.......$\mu F$.