$$\require{cancel}$$

# 2.2A: Point Charge

Let us arbitrarily assign the value zero to the potential at an infinite distance from a point charge $$Q$$. “The” potential at a distance $$r$$ from this charge is then the work required to move a unit positive charge from infinity to a distance $$r$$.

At a distance x from the charge, the field strength is $$\frac{Q}{4\pi\epsilon_0 x^2}$$. The work required to move a unit charge from $$x \text{ to }x + δx$$ is $$-\frac{Q\,\delta x}{4\pi\epsilon_0 x^2}$$. The work required to move unit charge from $$r$$ to infinity is $$-\frac{Q}{4\pi\epsilon_0}\int_r^{\infty}\frac{dx}{x^2}=-\frac{Q}{4\pi\epsilon_0 r}$$. The work required to move unit charge from infinity to $$r$$ is minus this.

Therefore

$V=+\frac{Q}{4\pi\epsilon_0 r}.\label{2.2.1}$

The mutual potential energy of two charges $$Q_1 \text{ and }Q_2$$ separated by a distance $$r$$ is the work required to bring them to this distance apart from an original infinite separation. This is

$P.E.=+\frac{Q_1Q_2}{4\pi\epsilon_0 r^2}\label{2.2.2}.$

Before proceeding, a little review is in order.

Field at a distance $$r$$ from a charge $$Q$$:

$E=\frac{Q}{4\pi\epsilon_0 r^2},\quad \quad \text{N C}^{-1} \text{ or } \text{V m}^{-1}$

or, in vector form,

$\textbf{E}=\frac{Q}{4\pi\epsilon_0 r^2}\hat{\textbf{r}}=\frac{Q}{4\pi\epsilon_0 r^3}\textbf{r}. \quad \quad \text{N C}^{-1}\text{ or }\text{V m}^{-1}$

Force between two charges, $$Q_1 \text{ and }Q_2$$:

$F=\frac{Q_1Q_2}{4\pi\epsilon r^2}.\quad \quad \text{N}$

Potential at a distance $$r$$ from a charge $$Q$$:

$V=\frac{Q}{4\pi\epsilon_0 r}.\quad \quad \text{V}$

Mutual potential energy between two charges:

$\text{P.E.}=\frac{Q_1Q_2}{4\pi\epsilon_0 r}.\quad \quad \text{J}$

We couldn’t possibly go wrong with any of these, could we?