1
$\begingroup$

Energy stored in the electric field or self energy of a solid sphere of radius $r$ and uniform charge distribution $q$ would be $$\frac{3}{5}\frac{1}{4\pi\epsilon_0}\frac{q^2}{r}$$

This result could be derived if we take electrostatic energy density $u = \frac{1}{2}\epsilon_0{E^2}$ where $E$ is the electric field intensity. From work-energy theorem, this energy would be the total work done to configure this uniformly charged solid sphere from infinity. Since the charge is quantised, $q=ne$ where $e$ is charge of an electron. If $dq=e$ quantised charged units are assembled from infinity, work done on the system gets stored in the electric field.

But when $n=1$ and if we consider it to be an electron with its classical radius, we get the value of self energy as $$\frac{3}{5}\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$$ Or at least we get something very finite.

I don't understand if this is the energy stored in the electric field and if so, is it intrinsic because with one $e$ charge, work-energy theorem doesn't make sense here. Or what about the error(s) in the assumptions leading to it?

Edit: I have come across this Wikipedia article (that I should have stumbled upon before asking here) and it happens to address most of my doubts (and inconsistencies).

Improve this question
$\endgroup$
5
  • $\begingroup$ $dq$ refers to any infinitesimal charge, not necessarily quantized charge $e$. It varies from system to system how small should $dq$ be. For an electron, you will have to assume $dq<<e$. For ordinary object $dq=e$ can be a good approximation. The point is dq must be much smaller than the total charge Q. $\endgroup$
    – KP99
    Commented Jul 23, 2021 at 8:41
  • $\begingroup$ @KP99 but if you quantise charge, $dq$ can only as small as $e$ $\endgroup$
    – Uphill trajectory
    Commented Jul 23, 2021 at 9:13
  • $\begingroup$ @KP99 i.e. I want to consider building the charge configuration from $e$ charges. Then the energy should should still come to $$\gamma\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$$ where $\gamma$ is some factor $\endgroup$
    – Uphill trajectory
    Commented Jul 23, 2021 at 9:22
  • $\begingroup$ Charge is quantized, but one doesn't invoke such condition when calculating self energy in classical regime (Coulombic interaction doesn't imply charge quantisation). More correct interpretation comes from quantum electro dynamics, where self interaction comes from an electron interacting with its own virtual photons. Here charge is quantized, and involves both relativistic and quantum mechanical corrections to Coulombic interaction $\endgroup$
    – KP99
    Commented Jul 23, 2021 at 10:20
  • $\begingroup$ @KP99 I understand your point now. Much thanks! $\endgroup$
    – Uphill trajectory
    Commented Jul 24, 2021 at 17:41

1 Answer 1

Reset to default
1
$\begingroup$

It's simply wrong!

An elementary particle or fundamental particle is a subatomic particle with no (currently known) substructure, i.e. it is not composed of other particles.

That means you can't consider an electron as a sphere of the finite radius with charge $e$. This is question is beyond the range of classical electrodynamics.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.