Derive an expression for the electric potential at a distance $r$ from a positive point charge $Q$.

(N/A) Consider a point charge $Q$ at the origin $O$. The electric potential at a point $P$ at a distance $r$ from $Q$ is defined as the work done by an external agent in bringing a unit positive test charge from infinity to point $P$.
Since the electrostatic force is conservative,the work done is independent of the path. We choose a radial path from infinity to $P$.
At an intermediate point $P^{\prime}$ at a distance $r^{\prime}$ from $O$,the electrostatic force on a unit positive test charge is:
$F = \frac{1}{4 \pi \epsilon_{0}} \frac{Q \times 1}{(r^{\prime})^{2}} = \frac{kQ}{(r^{\prime})^{2}}$
The work done by an external force $F_{ext}$ against the electrostatic force $F$ for a small displacement $dr^{\prime}$ towards the charge is:
$dW = F_{ext} \cdot dr^{\prime} = -F \cdot dr^{\prime} = -\frac{kQ}{(r^{\prime})^{2}} dr^{\prime}$
To find the total work done $W$ in bringing the charge from infinity to $r$,we integrate the expression:
$W = \int_{\infty}^{r} -\frac{kQ}{(r^{\prime})^{2}} dr^{\prime}$
$W = -kQ \int_{\infty}^{r} (r^{\prime})^{-2} dr^{\prime}$
$W = -kQ \left[ -\frac{1}{r^{\prime}} \right]_{\infty}^{r}$
$W = kQ \left[ \frac{1}{r} - \frac{1}{\infty} \right] = \frac{kQ}{r}$
Since electric potential $V = \frac{W}{q_{0}}$ and $q_{0} = 1 \text{ C}$,we have:
$V(r) = \frac{kQ}{r} = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r}$