Question

proved the guess law

Answer 1

Such surface is called Gauss' surface. This equation is called Gauss' law. Thus Gauss' law states that electric flux through any closed surface is equal to the net charge enclosed inside the surface divided by permittivity of vacuum. 

Answer 2

Gauss's law is the electrostatic equivalent of the divergence theorem. Charges are sources and sinks for electrostatic fields, so they are represented by the divergence of the field: ∇⋅E=ρϵ0∇⋅E=ρϵ0, where ρρ is charge density (this is the differential form of Gauss's law). You can derive this from Coulomb's law. For a charge density ρ(r⃗ ′)ρ(r→′),E(r⃗ )=14πϵ0∫Rρ(r⃗ ′)(r⃗ −r⃗ ′)ˆ∣∣r⃗ −r⃗ ′∣∣2dV=14πϵ0∫Rρ(r⃗ ′)(r⃗ −r⃗ ′)∣∣r⃗ −r⃗ ′∣∣3dV.E(r→)=14πϵ0∫Rρ(r→′)(r→−r→′)^|r→−r→′|2dV=14πϵ0∫Rρ(r→′)(r→−r→′)|r→−r→′|3dV.(This converges to E=q4πϵ0r2r^E=q4πϵ0r2r^ for a point charge.) Taking the divergence of this,∇⋅E=4π4πϵ0∫Rρ(r⃗ ′)δ(r⃗ −r⃗ ′)dV=ρ(r⃗ ′)ϵ0.∇⋅E=4π4πϵ0∫Rρ(r→′)δ(r→−r→′)dV=ρ(r→′)ϵ0.Then to arrive at the integral form presented in most introductory EM textbooks, the divergence theorem gives:∫V(∇⋅F⃗ )dV=∮∂VF⃗ ⋅dS→,∫V(∇⋅F→)dV=∮∂VF→⋅dS→,so in the case of electrostatics ∫Vρ(r⃗ )∫Vρ(r→) is simply all of the charge enclosed in VV so,∫V(∇⋅E⃗ )dV=∫Vρ(r⃗ )ϵ0dV=Qenclosed=∮∂VE⃗ ⋅dA→.