Proof of Gauss's law using coulomb's law

  • #1
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?
 

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  • #2
vela
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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 [itex]d\Omega[/itex] in terms of [itex]\vartheta[/itex] and [itex]\varphi[/itex] and what are the integration limits for the two variables?
 
  • #3
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Actually, you can use an arbitrary surface in this derivation.
d[tex]\Omega[/tex] is the solid angle subtended by the surface dS at the point where the charge q is placed.
[tex]\int[/tex]d[tex]\Omega[/tex] is the solid angle subtentded at the charge by the entire surface. it's value is 4[tex]\Pi[/tex]
 

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