# Energy of a Continuous Charge Distribution

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. \

• Adding charge densities and potentials does not make sense. You have one charge density as a function of position.
– nasu
Jun 19, 2022 at 15:26
• @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?
– JS30
Jun 19, 2022 at 15:44
• $$\rho = \rho_{1} + \rho_{2}$$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. Jun 19, 2022 at 15:54
• 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. Jun 19, 2022 at 16:26
• 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. Jun 19, 2022 at 17:38

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 $$V_{2}$$

Note:

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]