Two insulated metallic spheres of $3\,\mu F$ and $5\,\mu F$ capacitances are charged to $300\, V$ and $500\, V$ respectively. The energy loss,when they are connected by a wire,is (in $,J$)

$A$ parallel plate capacitor of capacitance $10 \mu F$ is charged by a $220 \text{ V}$ supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance $12 \mu F$. The loss of electrostatic energy in this process is (in $\text{ mJ}$)

In the following figure,$C_1=5 \mu F$,$C_2=C_3=10 \mu F$ and $\varepsilon=20 \text{ V}$. Initially,the switch $S$ is connected to point $A$ until capacitor $C_1$ is fully charged. Afterwards,the switch is thrown to the left side and connected to point $B$. The charge on capacitor $C_3$ after equilibrium is reached will be (in $\mu C$)

Two charged spheres of radii $a$ and $b$ are connected by a wire. The ratio of the electric fields on their surfaces $E_a/E_b$ is:

$A$ capacitor $C_{1}$ of capacitance $5\,\mu F$ is charged to a potential of $30\,V$ using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor $C_{2}$ of capacitance $10\,\mu F$ as shown in the figure. When the switch is closed,charge flows between the capacitors. At equilibrium,the charge on the capacitor $C_{2}$ is . . . . . . $\mu C$.

Two metal spheres have radii of $20\, cm$ and $10\, cm$ respectively, and each sphere carries a charge of $150\, \mu C$. After connecting them with a conducting wire, the common potential on them will be: