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

Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$,respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is $.....CV^2$.

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:

$A$ $10\,\mu F$ 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 $\dots \mu F$.

Two capacitors of $100 \mu F$ and $50 \mu F$ are connected in parallel. If the potential difference across $100 \mu F$ is $20 \text{ V}$ and across $50 \mu F$ is $40 \text{ V}$,then the common potential of the parallel combination will be (assuming same polarities of the capacitors are connected together).

$A$ capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of the resulting system: