In CGS: I know that:

$$ U_E = \frac12\sum_{i=1}^Nq_i\sum_{j\neq i}\frac{q_j}{|\mathbf{r}_i-\mathbf{r}_j|}. $$

In the continuous. Is it correct this conclussion?

$$ U_E=\frac12\int \rho(\mathbf{r})\varphi(\mathbf{r})d^3\mathbf{r}. $$

I know that this is not always true, because for just a lonely charge $q$ the energy is $U_E=0$, but with the formula this is $U_E\to\infty$. So, is the formula an aproximation or it's exact? we are adding to the integral the potential $\varphi_i(\mathbf r_i)$ generated by $q_i$, like this:

$$ U_E = \frac12\sum_{i=1}^Nq_i\sum_{j=1}^N\frac{q_j}{|\mathbf{r}_i-\mathbf{r}_j|}. $$

I suppose that when $\varphi,\rho$ are continous the formula is valid, bur I am not sure. I am confused because, this sounds to me a little artificial.


1 Answer 1


The equation of self-energy in the continuum limit is not valid for a point charge for a simple reason- While deriving the continuum limit, the charge density at each point is finite (ρ(r) in this case). But if you consider the charge density of a point charge, then it's infinite at the point where the charge is located. I remember Griffiths mentions this in his book- "Introduction to Electrodynamics" where he has devoted a whole section discussing this point. It's in section 2.4 of that book.


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