The total charge on a uniformly charged spherical shell having radius $R$ is $Q$. Then electric potential at a distance $r = R/2$ from the centre of the shell is . . . . . . .

There is a uniform electrostatic field in a region. The potential at various points on a small sphere centered at $P$,in the region,is found to vary between the limits $589.0\,V$ to $589.8\,V$. What is the potential at a point on the sphere whose radius vector makes an angle of $60^o$ with the direction of the field (in $,V$)?

Consider a sphere of radius $R$ having charge $q$ uniformly distributed inside it. At what minimum distance from its surface is the electric potential half of the electric potential at its centre?

$A$ spherical shell with an inner radius $a$ and an outer radius $b$ is made of conducting material. $A$ point charge $+Q$ is placed at the centre of the spherical shell and a total charge $-q$ is placed on the shell. Assume that the electrostatic potential is zero at an infinite distance from the spherical shell. The electrostatic potential at a distance $R$ $(a < R < b)$ from the centre of the shell is (where $K = \frac{1}{4\pi\varepsilon_0}$):

Electric potential at a point $P$ due to a point charge of $5 \times 10^{-9} \; C$ is $50 \; V$. The distance of $P$ from the point charge is ......... $cm$. (Assume,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \; N m^2 C^{-2}$)

The ratio of electric potentials at the point $E$ to that at the point $F$ is