Proof of Gauss's law using coulomb's law

[demonelite123]

so my book has a point charge q inside an arbitrary surface S and at a point P on the surface S, the electric field is E = q/4(π)(ε₀)(r²). so the flux through S is then dΦ = E * dS = q/4(π)(ε₀)(r²) * dS = q/4(π)(ε₀) * dScosθ /(r²) = q/4(π)(ε₀) * dΩ. then my book takes the integral of both sides and ends up with Φ = q/4(π)(ε₀) ∫ dΩ = q/4(π)(ε₀) (4π) = q/(ε₀).

what i don't understand it how did ∫ dΩ become 4π?

 

[vela]

It looks like they're not using an arbitrary surface. S is a sphere of radius r, and dS is a small element of area on the sphere. What's $d\Omega$ in terms of $\vartheta$ and $\varphi$ and what are the integration limits for the two variables?

 

[neeraj93]

Actually, you can use an arbitrary surface in this derivation.
d$$\Omega$$ is the solid angle subtended by the surface dS at the point where the charge q is placed.
$$\int$$d$$\Omega$$ is the solid angle subtentded at the charge by the entire surface. it's value is 4$$\Pi$$