Question

Define electric potential due to a point charge and arrive at the expression for the electric potential at a point due to a point charge in its vicinity.

Answer 1

The electrostatic potential at any point in an electrostatic field is defined as the work done in carrying a unit positive charge from infinity to that point against the electrostatic force of the field.

Consider a point Charge $+Q$ at $O$.
Let $P$ be a point at a distance $r$ from $O$.
Consider a point $A$ at a distance $x$ from $O$.
The magnitude of electrostatic force on unit-positive charge at $A$, is
$F=\frac{1}{4\pi {\epsilon }_0}\frac{Q\times i}{x^2}$
$F=\frac{Q}{4\pi {\epsilon }_0{x}^2}$ along $OP$
Work Done in moving a unit positive charge from $A$ to $B$ through a small distance $dx$ is,
$\Delta W=\overrightarrow{F}.d\overrightarrow{x}=Fdx\cos {180}^o$
$\Delta W=-Fdx$
$\Delta W=\frac{-Q}{4\pi {\epsilon }_0{x}^2}dx$
Total work done in moving a unit positive change from infinity to the point $P$ is,
$W={\int }_{\infty }^r\frac{-1}{4\pi {\epsilon }_0}\frac{Q}{x^2}dx=-\frac{Q}{4\pi {\epsilon }_0{x}^2}{\int }_{\infty }^r\frac{1}{x^2}dx$

$W=-\frac{Q}{4\pi {\epsilon }_0}{\left[\frac{1}{x}\right]}_{\infty }^r=\frac{Q}{4\pi {\epsilon }_0}[\frac{1}{r}-\frac{1}{\infty }]$

$W=\frac{Q}{4\pi {\epsilon }_{0}r}$
By definition, this work done is equal to the electrostatic potential (V) at that point due to $Q$.
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$