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?