Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

$A$ 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 . . . . . . .

$A$ $10\,\mu F$ capacitor is charged to a potential difference of $50\;V$ and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes $20\;V$. The capacitance of the second capacitor is....$\mu F$

Two capacitors $C_1$ and $C_2 = 2C_1$ are connected with a switch $S$ as shown in the figure. Initially,the switch is open and the charge on capacitor $C_1$ is $Q$. Now,when the switch is closed,the final charges on the capacitors are:

$A$ $600\,pF$ capacitor is charged by a $200\,V$ supply. It is then disconnected from the supply and connected to another uncharged $600\,pF$ capacitor. The electrostatic energy lost in the process is $.........\,\mu J$.

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is