Derive an expression for the electric potential in a electric field of positive point charge at distance $\mathrm{r}$.

Consider a point charge $Q$ at the origin as shown in figure.

The work done in bringing a unit positive test charge from infinity to the point $P$. For $Q$ which is positive the work done against the repulsive force on the test charge is positive. Since the work done is independent of the path we choose a convenient path along the radial direction from infinity to the point $P$.

At some intermediate point $P^{\prime}$ on the path, the electrostatic force on a unit positive charge is, $\mathrm{F}=\frac{k \mathrm{Q} \times 1}{\left(r^{\prime}\right)^{2}} \hat{r}^{\prime}$

$\ldots$ $(1)$

Where $\hat{r}^{\prime}$ is the unit vector along OP'. Work done against this force from $r^{\prime}$ to $r^{\prime}+\Delta r^{\prime}$ is, $\Delta \mathrm{W}=-\frac{k \mathrm{Q}}{\left(r^{\prime}\right)^{2}} \cdot \Delta r^{\prime} \quad \ldots$ $(2)$ $\quad[$ From $\mathrm{W}=\mathrm{F} r \cos \theta]$

Here, $\Delta r^{\prime}<0$ so, $\Delta \mathrm{W}>0$ in this formula.

Total work done by the external force is obtained by integrating above equation from $r^{\prime}=\infty$ to $r^{\prime}=r$.

$\therefore \mathrm{W}=-\int_{\infty}^{r} \frac{k \mathrm{Q}}{\left(r^{\prime}\right)^{2}} \cdot d r^{\prime} \quad\left[\lim _{\Delta r^{\prime} \rightarrow 0}=d r\right]$ $\therefore \mathrm{W}=-k \mathrm{Q}\left[-\frac{1}{r^{\prime}}\right]_{\infty}^{r}=\frac{k \mathrm{Q}}{r}$ $\therefore \mathrm{W}=\frac{\mathrm{Q}}{4 \pi \in_{0} r}$

From definition of electric potential,

$\mathrm{V}(r)=\frac{\mathrm{W}}{q}=\frac{\mathrm{W}}{1} \quad[\because q=1 \mathrm{C}]$

$\therefore \mathrm{V}(r)=\mathrm{W}=\frac{k \mathrm{Q}}{r}$