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Gauss's law derivation using solid angle

  1. Jul 17, 2011 #1
    So, I was trying to find a rigorous mathematical derivation of gauss's law(please I don't want to hear again any field lines nonsense) and I stumbled upon jackson's proof which uses the solid angle concept and seems a solid enough proof(stupid joke:smile:).The problem is that it's the first time I come in contact with solid angles and I don't get a certain part

    Eda=(q/4πε)(1/r^2)cosθda=(q/4πε)dΩ


    Why
    (1/r^2)cosθda=dΩ? (intuitively it makes sense but what is the actual proof)

    I'd appreciate if someone gave a clear from zero definition of solid angle

    Say we've proven Φ=Qen/ε
    if the charge lies outside of the surface how do you prove that the total flux is zero?
    Is there a solid angle approach or stoke's theorem is the easiest way?



    Forgive me for not using latex it's my first post
     
    Last edited: Jul 17, 2011
  2. jcsd
  3. Jul 17, 2011 #2

    Meir Achuz

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    There is a good derivation in Section 1.4 of Franklin "Classical Electromagnetism".
     
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