How to prove Gauss law geometrically (without coordinates) from Coulomb's law?

Question

Namely that the Flux of the electric field through any surface equals the charge enclosed/vacuum permittivity.

Without any coordinates or reference to axis.

Answer 1

If you start with Maxwell's equation you can apply the divergence theorem.

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ $$\int_V(\nabla \cdot \vec{E})dV = \frac{1}{\epsilon_0}\int_V \rho dV$$ $$\oint_A \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$$

Where $\oint_A \vec{E} \cdot d\vec{A}$ is the electric flux.

Answer 2

physic is proved by experiments, so enclose a charge in some surface and measure the flux, it is easier taking a conducting surface and measure the charge of the surface. this and of measurements finally came to create the Maxwell equations. and sometimes laws are explained by them. So you should say from what standpoint you want to proof the law.

Answer 3

This I think answers it. The point is just to show that the Flux through any surface is equal to the Flux through a sphere because any surface patch will always be (R/r) ^2 /cos¶ bigger that the corresponding sphere patch but it will be tilted by an angle ¶ reducing the Flux by cos¶. Since the electric field also drops as distance squared everything cancels out and you just get the electric field times the area of the sphere.