When two isolated conductors $A$ and $B$ are connected by a conducting wire,positive charge will flow from:

$A$ 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$)

There are two identical capacitors. The first one is uncharged and filled with a dielectric of constant $K$,while the second one is charged to potential $V$ and has air between its plates. If the two capacitors are joined in parallel,the common potential will be:

$A$ capacitor of $10\,\mu F$ charged up to $250\,V$ is connected in parallel with another capacitor of $5\,\mu F$ charged up to $100\,V$. The common potential is.....$V$.

$A$ $600\; pF$ capacitor is charged by a $200\; V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\; pF$ capacitor. How much electrostatic energy is lost in the process?

$A$ 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 . . . . . . .