$A$ 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 parallel plate capacitors are connected in series to a battery of $100\,V$. $A$ dielectric slab of dielectric constant $4.0$ is inserted between the plates of the second capacitor. The potential difference across the capacitors will now be respectively:

$A$ capacitor of capacitance $2 \mu F$ is charged with the help of a $60 \ V$ battery. After disconnecting the battery,if this capacitor is connected in parallel with another uncharged capacitor of capacitance $1 \mu F$,then the potential difference across the plates of the $2 \mu F$ capacitor is: (in $V$)

$A$ parallel plate capacitor of capacitance '$C$' is connected to a battery and charged to a potential '$V$'. Another capacitor of capacitance '$3C$' is charged to a potential '$3V$'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (in $CV^2$)

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

The energy of a charged capacitor is $U$. It is removed from the battery and then connected in parallel to another uncharged capacitor having capacitance twice that of the first one. The energy of the first and second capacitors respectively is . . . . . . .