On a similar note: when using Gauss' Law, do you even begin with Coulomb's law, or does one take it as given that flux is the surface integral of the Electric field in the direction of the normal to the surface at a point?
Answer
Let us for simplicity consider n point charges q1, …, qn, at positions →r1, …, →rn, in the electrostatic limit, with vacuum permittivity ϵ0.
Now let us sketch one possible strategy to prove Gauss' law from Coulomb's law:
Deduce from Coulomb's law that the electric field at position →r is →E(→r) = n∑i=1qi4πϵ0→r−→ri|→r−→ri|3.
Deduce the charge density ρ(→r) = n∑i=1qiδ3(→r−→ri).
Recall the following mathematical identity →∇⋅→r|→r|3 = 4πδ3(→r).
(This Phys.SE answer may be useful in proving eq.(3), which may also be written as ∇21|→r|=−4πδ3(→r)).Use eqs. (1)-(3) to prove Gauss' law in differential form →∇⋅→E = ρϵ0.
Deduce Gauss' law in integral form via the divergence theorem.
No comments:
Post a Comment