Electric Potential of a Point Charge

Let us calculate the electric potential $V({\bf r})$ generated by a point charge $q$ located at the origin. It is fairly obvious, by symmetry, and also by looking at Fig. 14, that $V$ is a function of $r$ only, where $r$ is the radial distance from the origin. Thus, without loss of generality, we can restrict our investigation to the potential $V(x)$ generated along the positive $x$-axis. The $x$-component of the electric field generated along this axis takes the form

$$E_x(x)= \frac{q}{4\pi\epsilon_0\,x^2}.$$ (94)

Both the $y$- and $z$-components of the field are zero. According to Eq. (87), $E_x(x)$ and $V(x)$ are related via

$$E_x(x) = - \frac{dV(x)}{dx}.$$ (95)

Thus, by integration,

$$V(x)= \frac{q}{4\pi\epsilon_0\,x} + V_0,$$ (96)

where $V_0$ is an arbitrary constant. Finally, making use of the fact that $V = V(r)$, we obtain

$$V({\bf r}) = \frac{q}{4\pi\epsilon_0\,r}.$$ (97)

Here, we have adopted the common convention that the potential at infinity is zero. A potential defined according to this convention is called an absolute potential.

Suppose that we have $N$ point charges distributed in space. Let the $i$th charge $q_i$ be located at position vector ${\bf r}_i$. Since electric potential is superposable, and is also a scalar quantity, the absolute potential at position vector ${\bf r}$ is simply the algebraic sum of the potentials generated by each charge taken in isolation:

$$V({\bf r}) = \sum_{i=1}^N \frac{q_i}{4\pi\epsilon_0\,\vert{\bf r} - {\bf r}_i\vert}.$$ (98)

The work $W$ we would perform in taking a charge $q$ from infinity and slowly moving it to point ${\bf r}$ is the same as the increase in electric potential energy of the charge during its journey [see Eq. (79)]. This, by definition, is equal to the product of the charge $q$ and the increase in the electric potential. This, finally, is the same as $q$ times the absolute potential at point ${\bf r}$: i.e.,

$$W = q\,V({\bf r}).$$ (99)