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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?

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  • $\begingroup$ Where am I wrong? Check-my-work questions are off-topic on this site. $\endgroup$
    – G. Smith
    Jun 24, 2020 at 17:49
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    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Jun 24, 2020 at 17:53

1 Answer 1

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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.

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