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Derive the Analog of Gauss' law

  1. Mar 29, 2016 #1
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    I am stuck with this (d) all the other (a),(b),(c) were just like
    (a) E= −∇φ
    (b) ∇ · E= −∇2φ
    (c) ∇ × E

    I know only very basic about Gauss' law, the problem is not from Electrodynamics course.
    I don't know what should I do to solve (d)
     
  2. jcsd
  3. Mar 29, 2016 #2
    Gauss' law is:
    [tex] \nabla \cdot E = \rho / \epsilon_0 [/tex]
    We also know that by definition, potential and electric field are related by
    [tex] E = - \nabla \cdot \phi [/tex]
    So we then have
    [tex] -\nabla^2 \phi = \rho/\epsilon_0 [/tex]

    You know the form of phi, so you can just integrate over r from 0 to R to get the charge contained in the region. Since there's no angular dependence, 4pi comes out, and it's just a simple integral over r.
    [tex] - 4\pi \epsilon_0 \int \limits_0^R (\nabla^2 \phi) r^2 dr = Q [/tex]

    I'll let you work out the rest.
     
  4. Mar 29, 2016 #3
    Thankyou!
     
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