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Proof of Coulomb's Law using Gauss' Law

  1. Jul 23, 2005 #1
    I'm trying to derive the vector field [itex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q\vec{r}}{r^3}[/itex] surrounding a point charge, starting with [itex]\oint_S \vec{E} \cdot \mathrm{d}\vec{A}[/itex]. My uneducated guess would be to get the magnitude of the electric field from gauss' law, then integrate to get the scalar potential, before taking the gradient to get the vector field. Is there a more elegant way to achieve this?


  2. jcsd
  3. Jul 23, 2005 #2


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    Use the freedom to choose any closed surface that encloses your charge to make your calculation easier. In the case of a point charge, it's easiest to choose a sphere (of radius r) centered at your point charge.

    Now, at each point on that sphere, observe that, by symmetry, the Electric Field must be radial and must have the same magnitude. In addition, the area elements [tex]d\vec A [/tex] are radially outward. Use these observations to sequentially simplify your integral.

    Since r is arbitrary (as long as your sphere encloses the point charge that it is centered on), your expression works for arbitrary r.

    (In a "proof" that involves the potential, you probably need to first prove that it exists. That is, show that the electric field is minus the gradient of a function for this situation. What condition must be imposed on the Electric Field?)
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