electric potential energy

The electric potential energy of a system of point charges is defined as the work required to assemble this system of charges by bringing them close together from an infinite distance (i.e. zero reference at \[r=\infty\]).

With the definition, we first derive the respective mathematical equation for the electric potential energy field of a single source charge.

The Coulomb force \[\mathbf{F}\] acting on a test charge \[q\] can be written in terms of the electric field \[\mathbf{E}\] as \[\mathbf{F}=q\mathbf{E}\]. The phrase "work required to assemble" tells us that we're doing work against the Coulomb force, so the resulting displacement when we move the charges represents a gain of potential energy. Note that The incremental work \[\Delta W\] done by moving the particle a short distance \[\Delta s\], assuming the change in \[\mathbf{F}\] is negligible, is then given as \[\Delta W\approx -\mathbf{F}\cdot \hat{\mathbf{s}}\Delta s\]. \[\hat{\mathbf{s}}\] is the unit vector in the direction of the motion and the negative indicates that the external force is against the field, i.e. \[\mathbf{F}_{\text{field}}=q\mathbf{E},\mathbf{F}_{\text{field}}=-\mathbf{F}_{\text{external}}\]. Substituting the equation for electric field at \[\mathbf{r}\], \[\Delta W\approx-q\mathbf{E}(\mathbf{r})\cdot \hat{\mathbf{s}}\Delta s\] gives us the work a very short distance around \[\mathbf{r}\].

With a basic framework we can now generalize the result. To get the work done over a larger distance, we need to take into account the possibility that \[\mathbf{E}\] varies along the path taken. One of the ways to do this is to sum contributions from points along the path traced out by the particle, i.e. \[W=\sum_{n=1}^{N}\Delta W(\mathbf{r}_{n})=\sum_{n=1}^{N}-q\mathbf{E}(\mathbf{r}_{n})\cdot \hat{\mathbf{s}}(\mathbf{r}_{n})\Delta s\]. Taking the limit \[\Delta s\to0\] we obtain \[W=-\int_{C}q\mathbf{E}(\mathbf{r})\cdot \hat{\mathbf{s}}(\mathbf{r})\,ds\], where \[C\] is the path followed (or the sequence of \[\mathbf{r}_{n}\]). Omitting the \[\mathbf{r}\] for clarity we get \[W=-\int_{C}q\mathbf{E}\cdot d\mathbf{s}\] where \[d\mathbf{s}=\hat{\mathbf{s}}\,ds\].

By definition, the change in electric potential energy, \[\Delta U\], of a point charge \[q\] that has moved from the reference position to position \[\mathbf{r}\] is \[U(\mathbf{r})-U(\infty)=\Delta U\]. Since electric force is conservative, the potential energy is also conservative, \[\Delta U=W=-\int_{\infty}^{r}q\mathbf{E}\cdot d\mathbf{s}\] (\[W=+\Delta U\] as we're doing work against the field). Define \[\mathbf{r}\] as the position vector of the test charge, \[r=\left\lVert \mathbf{r} \right\rVert\] as the radial distance from the source charge \[Q\] and \[d\mathbf{s}\] as the differential displacement vector along the chosen path. We also know that \[\mathbf{E}\] is parallel to \[\mathbf{s}\] as the electric field is radially directed from point charge \[Q\], thus:

\begin{align*} \Delta U&=-\int_{\infty}^{r}q\mathbf{E}\cdot d\mathbf{s}\\ &=-\int_{\infty}^{r}q \left\lVert \mathbf{E} \right\rVert \left\lVert d\mathbf{s} \right\rVert\cos(0^{\circ})\\ &=-\int_{\infty}^{r}qE\,ds\\ &=-\int_{\infty}^{r}q \left( \frac{1}{4\pi\epsilon_{0}}\frac{Q}{s^{2}} \right)\,ds\\ &=\biggl[ \frac{1}{4\pi\epsilon_{0}}\cdot \frac{Qq}{s} \biggr]^{r}_{\infty}\\ &=\frac{1}{4\pi\epsilon_{0}}\frac{Qq}{r} \end{align*}

Notice that the angle substituted in the dot product is 0 degrees.

We shall break down the directions of vectors and positive/negative signs for more clarity as to why it's zero degrees. First, we picked a radial path for integration (since any path is the same due to the conservative nature of EPE), so the differential displacement vector, \[d\mathbf{s}\] is also purely radial, i.e. \[d\mathbf{s}=dr\,\hat{\mathbf{r}}\]. This also means that \[d\mathbf{s}\] is always parallel to the \[\hat{\mathbf{r}}\] and the sign of \[dr\] determines whether the displacement is radially outward (\[dr>0\]) or radially inward (\[dr<0\]). Then, \[\mathbf{E}\] and \[d\mathbf{s}\] are directed along the radial unit vector \[\hat{\mathbf{r}}\], so they will be parallel to each other, thus the angle between them can only be \[0^{\circ}\] or \[180^{\circ}\].

Assuming the source charge \[Q\] is positive, \[\mathbf{E}\] points radially outward. If we were to move a test charge \[q\] from \[\infty\] towards the origin, \[dr<0\] and \[d\mathbf{s}\] points radially inward. Naturally, \[\mathbf{E}\] and \[d\mathbf{s}\] would be pointing in opposite directions, then it follows that the angle between them would be \[180^{\circ}\]. However, in the integral for potential energy, we are calculating the work done by the external force against the electric field. As defined above, if \[\mathbf{E}\] is radially outward, then \[\mathbf{F}_{\text{external}}\] (and \[d\mathbf{s}\]) must be radially inward. This tells us that \[\mathbf{F}_{\text{external}}\] and \[d\mathbf{s}\] have the same direction.

Reexamining the integral, \[\Delta U=-\int_{C}q\mathbf{E}\cdot d\mathbf{s}\], the negative sign in front already accounts for the work done against the field by the external force. So, we are only interested in the angle between \[\mathbf{E}\] and \[d\mathbf{s}\]. Recall that the magnitude of \[d\mathbf{s}\] is \[dr\]. If \[Q>0\], i.e. positive, \[\mathbf{E}\] is in the \[+\hat{\mathbf{r}}\] direction, then if we integrate from \[\infty\] to \[r\], \[dr\] is negative (thus \[d\mathbf{s}\] is pointing in the \[-\hat{\mathbf{r}}\] direction and \[\theta=180^{\circ}\]). However, when we write \[d\mathbf{s}=dr\cdot \hat{\mathbf{r}}\], \[dr\] itself already carries the sign of displacement. So, what happened inside the dot product was actually:

\begin{align*} \mathbf{E}\cdot d\mathbf{s}&=\left( \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}\hat{\mathbf{r}} \right)\cdot \left( dr\, \hat{\mathbf{r}} \right)\\ &=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}\left( \hat{\mathbf{r}}\cdot \hat{\mathbf{r}} \right)\,dr\\ \end{align*}

The dot product of two identical vectors is always 1 because the angle between them is obviously zero. Therefore, that is why in our derivation we substituted zero degrees. As with the directionality, regardless of the displacement being with or against the field, the sign of \[dr\] already handles it.