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

==== For r≥Rr\ge Rr≥R the radial electric field of the shell is ====

E(r)  =  Q4πε0 r2(unit: V/m).E(r) \;=\; \dfrac{Q}{4\pi\varepsilon_0\,r^2}\qquad(\text{unit: V/m}).E(r)=4πε0r2Q(unit: V/m).
Electromagnetic energy density (SI) is

u(r)  =  12 ε0E2(r)(unit: J/m3).u(r) \;=\; \tfrac12\,\varepsilon_0 E^2(r)\qquad(\text{unit: J/m}^3).u(r)=21ε0E2(r)(unit: J/m3).
Total EM energy stored outside the shell:

UEM=∫R∞12 ε0E2(r)  4πr2 dr=∫R∞12 ε0(Q4πε0r2)24πr2 dr.U_{\rm EM}
=\int_{R}^{\infty} \tfrac12\,\varepsilon_0 E^2(r)\; 4\pi r^2\,dr
= \int_R^\infty \tfrac12\,\varepsilon_0 \Big(\frac{Q}{4\pi\varepsilon_0 r^2}\Big)^2 4\pi r^2\,dr.UEM=∫R∞21ε0E2(r)4πr2dr=∫R∞21ε0(4πε0r2Q)24πr2dr.
Evaluate the integral:

UEM=Q28πε0∫R∞1r2 dr=Q28πε0R.U_{\rm EM}
= \frac{Q^2}{8\pi\varepsilon_0}\int_R^\infty \frac{1}{r^2}\,dr
= \frac{Q^2}{8\pi\varepsilon_0 R}.UEM=8πε0Q2∫R∞r21dr=8πε0RQ2.
(Units check: Q2/(4πε0R)Q^2/(4\pi\varepsilon_0 R)Q2/(4πε0R) has units J, so the factor is correct.)