- #1

demonelite123

- 219

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_{0})(r

^{2}). so the flux through S is then dΦ = E * dS = q/4(π)(ε

_{0})(r

^{2}) * dS = q/4(π)(ε

_{0}) * dScosθ /(r

^{2}) = q/4(π)(ε

_{0}) * dΩ. then my book takes the integral of both sides and ends up with Φ = q/4(π)(ε

_{0}) ∫ dΩ = q/4(π)(ε

_{0}) (4π) = q/(ε

_{0}).

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