The potential at a point, due to a positive charge of $100\,\mu C$ at a distance of $9\,m$, is

Derive an expression for electric potential at a point due to a system of $\mathrm{N}$ charges.

The radius of nucleus of silver (atomic number $=$ $47$) is $3.4 \times {10^{ - 14}}\,m$. The electric potential on the surface of nucleus is $(e = 1.6 \times {10^{ - 19}}\,C)$

Ten electrons are equally spaced and fixed around a circle of radius $R$. Relative to $V = 0$ at infinity, the electrostatic potential $V$ and the electric field $E$ at the centre $C$ are

A charge $+q$ is fixed at each of the points $x = x_0,\,x = 3x_0,\,x = 5x_0$, .... upto $\infty $ on $X-$  axis and charge $-q$ is fixed on each of the points $x = 2x_0,\,x = 4x_0,\,x = 6x_0$, .... upto $\infty $ . Here $x_0$ is a positive constant. Take the potential at a point due to a charge $Q$ at a distance $r$ from it to be $\frac{Q}{{4\pi {\varepsilon _0}r}}$. Then the potential at the origin due to above system of charges will be

At distance of $5$ $cm$ and $10$ $cm $ outwards from the surface of a uniformly charged solid sphere, the potentials are $100$ $V$ and $75$ $V$ respectively . Then