Editing Openai/6897769e-4ee4-800f-aba5-69cca34f701c (section)

=== Electromagnetic self-energy of the thin shell: ===

UEM=Q28πε0R(we used this before).U_{\rm EM}=\frac{Q^2}{8\pi\varepsilon_0 R} \quad(\text{we used this before}).UEM=8πε0RQ2(we used this before).
For the electron-scale shell UEMU_{\rm EM}UEM equals the electron rest energy mec2≈8.187×10−14 Jm_e c^2 \approx 8.187\times10^{-14}\ \mathrm{J}mec2≈8.187×10−14 J.

Surface (tension) energy stored in the membrane is

Etens  =  T⋅(area)  =  T⋅4πR2.E_{\rm tens} \;=\; T \cdot (\text{area}) \;=\; T \cdot 4\pi R^{2}.Etens=T⋅(area)=T⋅4πR2.
With the equilibrium tension above, one finds (direct substitution) the simple relation

Etens  =  12 UEM.E_{\rm tens} \;=\; \frac{1}{2}\,U_{\rm EM}.Etens=21UEM.
Numerically for the electron:
* UEM≈8.187×10−14 JU_{\rm EM}\approx 8.187\times10^{-14}\ \mathrm{J}UEM≈8.187×10−14 J
* Etens≈4.094×10−14 JE_{\rm tens}\approx 4.094\times10^{-14}\ \mathrm{J}Etens≈4.094×10−14 J

So the tension (binding) energy is exactly half of the total EM self-energy — the other half is the field energy. In the thin-shell classical equilibrium model the binding stress contributes ~50% of the stored energy. That’s why earlier we saw the factor 1/21/21/2 showing up: the shell model partitions energy half/half between field and structural (binding) energy when in equilibrium.