# Gauss's law derivation using solid angle

1. Jul 17, 2011

### Sionsd

So, I was trying to find a rigorous mathematical derivation of gauss's law(please I don't want to hear again any field lines nonsense) and I stumbled upon jackson's proof which uses the solid angle concept and seems a solid enough proof(stupid joke).The problem is that it's the first time I come in contact with solid angles and I don't get a certain part

Eda=(q/4πε)(1/r^2)cosθda=(q/4πε)dΩ

Why
(1/r^2)cosθda=dΩ? (intuitively it makes sense but what is the actual proof)

I'd appreciate if someone gave a clear from zero definition of solid angle

Say we've proven Φ=Qen/ε
if the charge lies outside of the surface how do you prove that the total flux is zero?
Is there a solid angle approach or stoke's theorem is the easiest way?

Forgive me for not using latex it's my first post

Last edited: Jul 17, 2011
2. Jul 17, 2011

### Meir Achuz

There is a good derivation in Section 1.4 of Franklin "Classical Electromagnetism".