B Distribution of energy in the electric field surrounding an electron

[Herbascious J]

TL;DR Summary

How is the energy distributed geometrically/spacially in the electric field surrounding an electron?

I am thinking about how an electric field has energy associated with it. If a single electron exists alone in a remote vaccuum, I believe it has it's own electric field surrounding it, and that this field has an energy content associated with it. My question is; does this electric field store energy in it in a geometrical/spacial way? for example, it I 'slice up' the space around the electron into concentric shells, like an onion, is the energy of the electric field stored in each concentric shell in some mathematical way? Is there a gradient in each shell where there is less and less energy and therefore less and less overall electric field by volume somehow? I am specifically trying to understand the distribution of the energy through out the field in a geometrical/spacial way. Any interpretation of this question is welcome. Thank you!

 

[kuruman]

Energy is distributed in the space surrounding the electron. The energy density function (energy per unit volume) is given by $u=\frac{1}{2}\epsilon_0E^2$ where $E$ is the magnitude of the electric field. It is given by $E=\dfrac{e}{4\pi\epsilon_0r^2}~$where $e$ is the charge of the electron. For example, the electrostatic energy contained within a shell of inner radius $R_1$ and outer radius $R_2$ with the electron at its center is given by $$U=\int u~dV=\frac{1}{2}\epsilon_0\int_{R_1}^{R_2}\left( \frac{e}{4\pi\epsilon_0r^2}\right)^24\pi r^2~dr=\frac{e^2}{8\pi\epsilon_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$Note that the energy density is not uniform but depends on the distance $r$ from the center and decreases as one goes farther from it.

 

[kuruman]

Thank you for the formula. I apologize for my weakness with calculus...the way I'm understanding it is that the energy density (energy/volume, I'm assuming this is like an average) is defined by the distance from the source. So, a shell with outer radius 10 and inner radius 9 would have an average density proportional to [1/10 - 1/9] = 0.011. Alternatively the volume from R=0 to R=1 would have an average density proportional to 1. Again I apologize for being a bit choppy with the math, but is it safe to say that the density is dropping off by the invesrse of the square of the distance? For example, if I move away from the source by 10x, then the density of energy in that region is 100x less dense, etc?

It is better to think of energy density as analogous to mass density not as an average. If the density is uniform, i.e. the same everywhere, then you can say that mass (or energy) is density times volume. If it is not uniform, then you have to use calculus.

If you must have an average density, then you have to calculate it correctly. The average density is given by

$u_{\text{avg.}}=\dfrac{\text{Total energy contained in a volume}}{\text{The volume}}.~$In this example the average density would be $$u_{\text{avg.}}=\frac{\frac{e^2}{8\pi\epsilon_0}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}{\frac{4}{3}\pi(R_2^3-R_1^3)}.$$ Also note that you cannot set $R_1$ equal to zero, because it is in the denominator and $\dfrac{1}{0}$ is undefined, it is not zero as you seem to think.

Finally, the energy density goes as the inverse fourth power of the distance from the center ($\sim\frac{1}{r^4}$) because the electric field is inversely proportional to the square of the distance and the density is proportional to the field squared.