# 2.2A: Point Charge

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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?