Define electrostatic potential and its unit. Obtain expression for electrostatic potential at a point P in the field due to a point charge.

Hint: To answer this question, we have to use the concept of the potential energy for a system of two charges. One of the two charges should be taken as unit positive charge.

Formula Used:
 $\Rightarrow {F_E} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}} $ , where $ {F_E} $ is the electrostatic force between the two charges $ Q $ and $ q $ separated by a distance $ x $.

Complete step by step answer:
The electrostatic potential at a point in an electric field is defined as the work done to bring a unit positive charge from infinity to that point without acceleration. For determining the unit of the electrostatic potential, we focus on its definition. It is the work done per unit charge. We know that the SI unit of work is Joule, and that of the charge is Coulomb. So, the unit of electrostatic potential becomes Joule per Coulomb, or $ J/C $.
Let a point charge $ Q $ be fixed at a point A. We consider the electrostatic potential at the point B, at a distance $ r $ from the point A, as shown below.

Now, as per the definition of the electrostatic potential, we need to bring a unit positive test charge $ q $ from infinity to the point B without acceleration. The work done in this process will be equal to the electrostatic potential at point B. There are two forces are acting on $ q $:
- Electrostatic force due to the point charge $ Q $ $ \left( {{F_E}} \right) $
- Force applied externally $ \left( F \right) $
Since the test charge should not have acceleration, the net force acting on it must be equal to zero. So
 $\Rightarrow F = - {F_E} $ (i)
For the test charge $ q $ at a distance of $ x $ from the point A, the electrostatic force on $ q $ is given by the Coulomb’s law as
 $\Rightarrow {F_E} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}} $
From (i)
 $\Rightarrow F = - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}} $ (ii)
Now, work done in displacing the test charge by $ dx $
 $\Rightarrow dW = F \cdot dx = Fdx\cos \theta $
Since the force $ F $ is parallel to the displacement $ dx $ , so $ \theta = 0 $ .
 $ \therefore dW = Fdx $
From (ii)
 $\Rightarrow dW = - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}}dx $
As the test charge is brought from infinity to point A, $ x $ varies from $ \infty $ to $ r $ . Integrating between these limits, we get
 $\Rightarrow \int_0^W {dW} = \int_\infty ^r { - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Qq}}{{{x^2}}}dx} $
Taking the constant outside the integral
 $\Rightarrow \int_0^W {dW} = - \dfrac{{Qq}}{{4\pi {\varepsilon _0}}}\int_\infty ^r {\dfrac{{dx}}{{{x^2}}}} $
 $\Rightarrow \left[ W \right]_0^W = - \dfrac{{Qq}}{{4\pi {\varepsilon _0}}}\left[ { - \dfrac{1}{x}} \right]_\infty ^r $
Substituting the limits, we get
 $\Rightarrow W = - \dfrac{{Qq}}{{4\pi {\varepsilon _0}}}\left( { - \dfrac{1}{r}} \right) $
 $\Rightarrow W = \dfrac{{Qq}}{{4\pi {\varepsilon _0}r}} $
As the test charge $ q $ is a unit positive charge, so $ q = + 1C $ .