# Proof of Gauss's law using coulomb's law

1. Mar 1, 2010

### 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?

2. Mar 2, 2010

### vela

Staff Emeritus
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?

3. Mar 2, 2010

### 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$$\Pi$$