Where Does Feynman Get This Equation For E Field?

[Hornbein]

http://www.feynmanlectures.caltech.edu/I_28.html#mjx-eqn-EqI283

It's Coulomb's law with retarded positions for the charges. Or something like that. How is it derived, please?

Oh, never mind. I found some nice instructive materials from U Texas.

 

[gtoguy97]

The derivation of Coulomb's law with retarded positions is actually quite simple. It is based on the fact that the electric field at a point in space is generated by all charges in the universe. If we assume that the charges are moving at a constant speed, then the contribution to the field from any given charge must be calculated using the position of the charge at an earlier time, which is known as the "retarded position".To start the derivation, we can use Gauss's law, which states that the net electric flux through any closed surface is equal to the enclosed charge divided by the permittivity of free space:∮E⋅dA = q/ε0where E is the electric field, A is the area of the surface, ε0 is the permittivity of free space, and q is the charge enclosed by the surface.We can rewrite this equation in terms of the electric potential V, which is related to the electric field by the relationE = -∇Vwhere ∇ is the gradient operator. This gives us∫∇V⋅dA = q/ε0where the integral is taken over a closed surface enclosing the charge q.Now, we can apply the principle of superposition to this equation. That is, we can add together the contributions to the electric potential from each individual charge in the universe, weighted by its retarded position. This gives us∫∇[∑q r(t-t')/4πε0 r2]⋅dA = q/ε0where q is the charge of the individual particle, r(t-t') is the retarded position of the charge at time t', and r is the distance from the charge to the point of observation.We can simplify this equation further by noting that the gradient of the retarded position is zero outside of the light cone, since the charge can only affect the electric potential at points within the light cone. Thus, we can rewrite the equation as∫∇[∑q δ(t-t')/4πε0 r]⋅dA = q/ε0where δ(t-t') is the Dirac delta function.Finally, we can solve this equation to get