I had a question regarding the derivation for the following expression of the energy of a continuous charge distribution $$W=\frac{\epsilon_0}{2}\int_\text{all space} E^2d\tau$$ To get this result, we start from the fact that when considering point charges, the work is given by the sum $$W=\frac{1}{2}\sum_{i=1}^nq_1V(\vec{r}_i)=\frac{1}{2}\sum_{n=1}^{n}\sum_{\begin{align*}j=1\\j\ne i\end{align*}}^n\frac {q_iq_j}{4\pi\epsilon_0}\frac{1}{r_{ij}}$$ Then, we extend this to continuous distributions by making it a volume integral, and taking into account the charge distributions in two regions of space $$W=\frac{1}{2}\int_\text{all space}\frac{\rho_1\rho_2}{4\pi\epsilon_0}\frac{1}{r_{12}}d\tau_1d\tau_2$$ We can see that $\int\frac{\rho_2}{4\pi\epsilon_0}\frac{1}{r_{ij}}d\tau_2$ is just the potential at region 1, due to the charge distribution in region 2. Hence we can rewite this as $$W=\frac{1}{2}\int_\text{all space}\rho Vd\tau$$ I understand that $\rho=\rho_1+\rho_2$, and that $V=V_1+V_2$, then $$\frac{1}{2}\int\rho Vd\tau=\frac{1}{2}(\int\rho_1 V_1d\tau+\int\rho_1V_2d\tau+\int\rho_2V_1d\tau+\int\rho_2V_2d\tau)$$ Where the two integrals in the middle are equal, so by dividing their sum by two we get the total work. However, the terms which include the product between a potential and its own charge distribution should vanish, yet I haven't been able to see how this happens, since when I try to solve those integrals, like $\int\rho_1V_1d\tau$, the result diverges. I wanted to know whether $\rho$ and $V$ are what I understand them to be, and if so how does the integral vanish, or If this is wrong, then what charge distribution and potential do $\rho$ and $V$ stand for. \

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    $\begingroup$ Adding charge densities and potentials does not make sense. You have one charge density as a function of position. $\endgroup$
    – nasu
    Jun 19, 2022 at 15:26
  • $\begingroup$ @nasu So are you saying that \rho contains the information of both \rho_1 and \rho_2 and so does V with V_1 and V_2? How do we construct \rho from the original distributions and V from the original potentials, and also if \rho accounts for the distribution of charge in all space and V for the potential in all space as well, then aren't we counting the energy of a charge distribution due to its own potential too? $\endgroup$
    – JS30
    Jun 19, 2022 at 15:44
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    $\begingroup$ $$\rho = \rho_{1} + \rho_{2}$$does make sense. It is simply a single charge density function split into 2 distinct elements. One from one e.g sphere and one from another. $\endgroup$ Jun 19, 2022 at 15:54
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    $\begingroup$ The total energy of 2 seperate charge distributions, is not.just the potential energy between them. You need to also include the work required to build.up the individual charges in the presence of itself. $\endgroup$ Jun 19, 2022 at 16:26
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    $\begingroup$ You are using symbol $W$ for two different energies: 1) electrostatic energy of point charges 2) electrostatic energy of a continuous charge distribution. These are not the same energies because the system is different. Neither is 1) limit of 2) when charge is continuously concentrated into points. $\endgroup$ Jun 19, 2022 at 17:38

1 Answer 1


You will soon see that the splitting of charge density and potential into 2 distinct elements, is the same as splitting E into 2 elements.

$$\vec{E}_{total} = \vec{E}_{1} + \vec{E}_{2}$$

$$W = \frac{1}{2} \epsilon_{0} \iiint |\vec{E}_{total}|^2 d^3 r$$

$$W = \frac{1}{2} \epsilon_{0} \iiint |\vec{E}_{1} + \vec{E}_{2}|^2 d^3 r$$

Computing this expression gives us 3 distinct terms. $$W = \frac{1}{2} \epsilon_{0} \iiint |\vec{E}_{1} |^2 d^3 r $$

$$+\frac{1}{2} \epsilon_{0} \iiint |\vec{E}_{2} |^2 d^3 r $$ $$+\epsilon_{0} \iiint \vec{E}_{1}\cdot \vec{E}_{2} d^3 r $$

What do they represent?

The first term represents the energy of $\vec{E}_{1}$

The second term represents the energy of $\vec{E}_{2}$

The third term represents the potential energy between the charge distributions (building up field 1 in the presence of field 2), I'll leave it to you to prove this!

Splitting up charge density:

This decomposition of E into 2 elements.is the same as splitting up the charge distribution into 2 elements

$$\rho = \rho_{1} + \rho_{2}$$

$$V = V_{1} + V_{2}$$

$V_{1}$ is caused by $\rho_{1}$, and $V_{2}$ is caused by $\rho_{2}$

$$ W= \frac{1}{2}\iiint [\rho_{1} + \rho_{2}][V_{1} + V_{2}] d^3r$$

There are 3 distinct terms of this expression

$$W= \frac{1}{2}\iiint \rho_{1}V_{1} d^3r$$

$$+\frac{1}{2}\iiint \rho_{2}V_{2} d^3r$$

$$+\frac{1}{2}\iiint [\rho_{1}V_{2} + \rho_{2} V_{1}] d^3r$$

The first 2 terms take the form that we are familiar with, they do not vanish. They are the individual energies of $\vec{E}_{1}$ and $\vec{E}_{2}$.

The last term is slightly more complicated.

This term represents the potential energy between the 2 charge distributions!

To show this:

$$\iiint \rho_{1}V_{2} d^3r = \iiint \rho_{2}V_{1} d^3r$$

As building up distribution 1 in the presence of potential 2, is the same as building up distribution 2 in the presence of potential 1 [which is intuitive, you can also prove this mathematically]

Substituting this identity into our third term, reveals that this term. . Is infact

$$\iiint \rho_{1} V_{2} d^3r$$

This is obviously the potential energy between our charge distributions since we are building up a charge $\rho_{1} d^3r$ in the presence of $V_{2}$


You say the first 2 terms diverge, if your using this expression for a point charge then yes, the field energy is infinite, if you use these formulas. This formula is not valid for point charges since the derivation assumes $\rho$ is finite[discussed further in griffiths]. Instead we model energy of point charges using the discrete formula you mentioned, or using renormalisation]


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