# Radius of electron and electrostatic energy [closed]

The classical radius of electron is obtained from the electrostatic energy of a sphere of radius $$r_e$$ which is: $$U=\displaystyle\frac{e^2}{4\pi \epsilon_0r_e}$$. For the electrostatic energy of a sphere of radius $$R$$ I obtain that $$U=\displaystyle\frac{3e^2}{20\pi\epsilon_0R}$$ by using the formula $$U=\displaystyle\frac{\epsilon_0}{2}\int{E^2d\tau}$$. Where am I wrong?

• Where am I wrong? Check-my-work questions are off-topic on this site. Commented Jun 24, 2020 at 17:49
• If check-my-work questions were allowed, we would have to actually see your work to check it. All you have shown is the final result, not how you got it. Commented Jun 24, 2020 at 17:53

The classical electron radius is only a very rough approximation. $$r_{e}$$ just tells you the order of magnitude radius at which the electrostatic energy of a charged sphere is $$mc^{2}$$. It is not really possible to do better than an order-of-magnitude estimate anyway, because the precise value of the energy depends on how the charge is distributed over the sphere. For example, if the charge is all located at radius $$R$$, the energy is $$U=\frac{e^{2}}{8\pi\epsilon_{0}R}$$. With different arrangements of the charge density internal to $$r=R$$, the potential energy required to confine the charge in that region can be pretty much anything larger than that. (We know that having the charge uniformly distributed at $$r=R$$ is the lowest-energy configuration, because that is the way the charge would arrange itself on a conducting sphere of equal radius.)
In reality, the electron is known to be pointlike down to a scale much smaller than $$r_{e}$$, so the precise value of $$r_{e}$$ has no real physical meaning. $$r_{e}$$ is just a useful way to parameterize the electrostatic energy of a small charge, and it appears in the cross-section for electron-photon scattering.