Question

Derive an expression for the electric potential due to a point charge :

Answer 1

Consider the electric potential due to a point charge q, As we move from point A, at distance r_A from the charge q, to point B, at distance r_B from the charge q, the change in electric potential is

$\Delta V_{BA}=V_B-V_A=-{\int }_A^{B}E.ds$

$E.ds=\left[k\frac{q}{r^2}\right]\hat{r}.ds$

$\hat{r}.ds=dr$

Only the radial distance r determines the work done or the potential. We can move through any angle we like and, as long as the radial distance remains constant, no work is done or there is no change in the electric potential.

$\Delta V_{BA}=V_B-V_A=-{\int }_{r_A}^{r_B}Edr=-{\int }_{r_A}^{r_B}\left[k\frac{q}{r^2}dr\right]$

$\Delta V_{BA}=V_B-V_A=-kq\left[\right(-1)r^{-1}{]}_{r_A}^{r_B}$

$\Delta V_{BA}=V_B-V_A=kq[\frac{1}{r_B}-\frac{1}{r_B}]$

This is the change in electric potential due to a point charge as we move from r_A to r_B.

We could ask about the change in electric potential energy as we move a charge q' from radius r_A to r_B due to a point charge q

$\Delta U_{BA}=k{q}^{\prime }q[\frac{1}{r_B}-\frac{1}{r_A}]$

As with gravitational potential energy, it is more convenient -- and, therefore, useful -- to talk about the electric potential energy or the electric potential relative to some reference point. We will choose that reference point to be infinity. That is,

$r_A=\infty$

That means we can then write the electric potential at some radius r as $V=kq\frac{1}{r}$