This questions comes from reading through the fourth edition of Griffith's Introduction to Electrodynamics.

Consider a point charge q located at the origin. Its potential is given exactly by $V_0 = \frac1{4\pi\epsilon_0}\frac{q}{r}$. Now, consider the same point charge q located a distance d along the +x axis. Its potential is approximated by $V_d = \frac1{4\pi\epsilon_0}(\frac{q}{r}+\frac{dq}{r^2})$ (via multipole expansion).

It seems that the energies (computed by $\int{V}{\rho}{d\tau}$ (where $\rho$ is charge density; $d\tau$ is infinitesimal volume; $V$ is potential)) will be different between the two situations. However, I am finding this confusing and difficult to believe.

  1. I thought one way to compute the energy of a charge distribution was to imagine bringing in the charge from infinity away and computing how much work is needed to do so (assuming charges don't accelerate to get there, etc.). This analysis obviously gives you 0 work (you are bringing a single charge to a location in vacuum; thus, no external electric field doing work on the charge coming into its location). Why doesn't this analysis apply for this situation?

  2. Are the energies in the situations actually different? If we interpret that energy is contained in the electric field, why would the electric field produced by a charge at the origin have different energy than the electric field produced by the same charge nudged over a bit along the +x axis?


1 Answer 1


Calculating the energy for the point charge at the origin, or displaced to some other point will give the same result. If you're getting a different answer, then you should show your calculations; you've definitely made an error.

For your first question, the answer of $0$ corresponds to the work you need to do to move a charge from infinity and place it at the origin. This is indeed $0$. However, a slightly different quantity to calculate is the total energy of the entire configuration (including the energy due to the electric field at the position of the charge itself). This will yield $\infty$. These two answers are not contradictory, but rather they're answering different questions. I'm pretty sure Griffiths has a discussion about this in that section of the text. The point charge's integral $\frac{1}{2}\int \rho V\,d\tau$, or equivalently $\int\frac{\epsilon_0}{2}\|\mathbf{E}\|^2\,d\tau$ being infinite for a point charge follows because of the $\frac{1}{r^2}$ decay in the Coulomb solution for the field: \begin{align} \int_{\Bbb{R}^3}\|\mathbf{E}\|^2\,d\tau\propto \int_0^r\left(\frac{1}{r^2}\right)^2\,4\pi r^2\,dr\propto \int_0^{\infty}\frac{1}{r^2}\,dr=\infty. \end{align}

On the other hand, if you calculate this integral for a uniformly distributed solid ball of charge of a fixed radius $R$, then you'll get some finite value, because for $r>R$, the electric field behaves as $\frac{1}{r^2}$, so integrating the field in this outer region gives you something proportional to $\int_R^{\infty}\frac{1}{r^2}\,dr=\frac{1}{R}$, which is finite; next for the region $r\leq R$, the electric field grows linearly $\mathbf{E}\propto r\,\hat{\mathbf{r}}$, so the integral in this inner region is clearly finite, hence to full integral over all of $\Bbb{R}^3$ is also finite. The problem of infinite energy of electron as point charge? talks about this issue.

Once again to emphasize: the above results do not depend on the location of the point charge. You can make a simple translation in the integration so that you can re-center everything to the origin (if you want to be really pedantic, we're exploiting the translation-invariance of Lebesgue measure on $\Bbb{R}^n$). Therefore, if you get a different answer in the case of a non-origin point charge, you should double check the computations.


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