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,$\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \; N m^2 C^{-2}$)

$A$ particle $A$ has charge $+q$ and particle $B$ has charge $+4q$,with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference $V$,the ratio of their speeds $\frac{v_A}{v_B}$ will be:

For a hollow spherical shell,how does the potential $(V)$ change with respect to the distance $(r)$ from the centre?

$A$ charge $q$ is placed at the centre $O$ of a circle of radius $R$. Two other charges $q$ and $q$ are placed at the ends of the diameter $AB$ of the circle. The work done to move the charge at point $B$ along the circumference of the circle to a point $C$ as shown in the figure is

The unit of the physical quantity obtained by the line integral of the electric field is . . . . . . .

The electrostatic potential on the surface of a uniformly charged spherical shell of radius $R = 10 \ cm$ is $120 \ V$. The potential at the centre of the shell,at a distance $r = 5 \ cm$ from the centre,and at a distance $r = 15 \ cm$ from the centre of the shell respectively,are: