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:

$A$ $4 \;\mu F$ capacitor is charged by a $200 \;V$ supply. It is then disconnected from the supply and connected to another uncharged $2 \;\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

If the capacitor connected across $AB$ is charged to $Q$ and then the switch is shifted to position $2$,calculate the heat released after that.

The energy stored in the capacitor shown in figure $(a)$ is $4.5 \times 10^{-6} \ J$. If the battery is disconnected and the capacitor is connected to another uncharged capacitor of $900 \ pF$ as shown in figure $(b)$,what will be the total energy of the system?

$A$ conducting sphere $S_1$ of radius $r_1 = 3 \text{ cm}$ is connected by a conducting wire to another conducting sphere $S_2$ of radius $r_2 = 2 \text{ cm}$. Before they are connected,$S_1$ carries a charge of $10 \text{ units}$. The electric potential at a point which is at a distance $4 \text{ cm}$ from the centre of $S_1$ and a distance $3 \text{ cm}$ from the centre of $S_2$ is:

$A$ 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 $\text{ mJ}$)