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Proof of Gauss's law, starting from Coulomb's Law?

  1. Jan 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Provide a proof of Gauss’s law starting from Coulomb’s law for a point charge


    2. Relevant equations
    F=kq1.q2/r^2

    (closed)s∫E.n.da=qenc/ε0


    3. The attempt at a solution
    My starting point is:

    F21=1/(4∏.ε0).(q2.q1)/r21^2.r(unit vector)21

    *Note, the 21 terms after F and r should be subscripts

    and here's where I've ended up:

    E*vector*(r*vector*)=[F*vector*(r*vector*)]/q0= 1/(4∏.ε0).Ʃi(qi)/ri0^2.r(unit vector)i0

    Not sure if I'm even on the right track. If anyone could provide the proof it'd be much appreciated.
     
  2. jcsd
  3. Jan 1, 2013 #2
    Coulomb's law for a point charge in the origin is:

    [itex]\mathbf{E}\left( \mathbf{r} \right)=\frac{q}{4\pi {{\varepsilon }_{0}}}\frac{{\mathbf{\hat{r}}}}{{{r}^{2}}}[/itex]

    When you integrate the field on an arbitrary closed surface (that includes the origin):

    [itex]\oint\limits_{S}{\mathbf{E}\cdot d\mathbf{S}}=\frac{q}{4\pi {{\varepsilon }_{0}}}\oint\limits_{S}{\frac{1}{{{r}^{2}}}\mathbf{\hat{r}}\cdot d\mathbf{S}}[/itex]

    because of the inner product [itex]\mathbf{\hat{r}}\cdot d\mathbf{S}[/itex] , the surface element [itex]d\mathbf{S}[/itex] projects onto a spherical surface of radius [itex]r[/itex] , so the surface integral is equivalent to:

    [itex]\oint\limits_{S}{\frac{1}{{{r}^{2}}}\mathbf{\hat{r}}\cdot d\mathbf{S}}=\int_{\theta =0}^{\pi }{\int_{\varphi =0}^{2\pi }{\frac{1}{{{r}^{2}}}{{r}^{2}}\sin \theta d\varphi d\theta }}=4\pi [/itex]

    Alternatively, you could start from the delta function formula:

    [itex]\nabla \cdot \left( \frac{{\mathbf{\hat{r}}}}{4\pi {{r}^{2}}} \right)=\delta \left( \mathbf{r} \right) [/itex]

    to evaluate the divergence of the electric field:

    [itex]\nabla \cdot \mathbf{E}\left( \mathbf{r} \right)=\frac{q}{{{\varepsilon }_{0}}}\delta \left( \mathbf{r} \right) [/itex]

    Then, applying the divergence theorem in the volume integral of [itex]\nabla \cdot \mathbf{E}\left( \mathbf{r} \right) [/itex] , you get:

    [itex]\oint\limits_{S=\partial V}{\mathbf{E}\cdot d\mathbf{S}}=\int\limits_{V}{\nabla \cdot \mathbf{E}dV}=\int\limits_{V}{\frac{q}{{{\varepsilon }_{0}}}\delta \left( \mathbf{r} \right)dV}=\frac{q}{{{\varepsilon }_{0}}}[/itex]
     
  4. Jan 5, 2013 #3

    rude man

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    Homework Helper
    Gold Member

    Easy way is to assume Gauss' law is right, and derive Coulomb's law from it.
     
  5. Jan 7, 2013 #4
  6. Jan 7, 2013 #5

    BruceW

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    Homework Helper

    Gauss' law is more general than Coulomb's law, because in Coulomb's law, we assume stationary charges. So, for the title of this thread, I think it should be: "Show that if Coulomb's law holds, then so does Gauss' law". Anyway, sorry to be nitpicky.

    Dez1, since this is the homework section, we are meant to offer help, but not to give you the answer. It will be better practice if you try to have a good attempt at the problem yourself. Your working is looking good so far. Your final line of working (if I read it right) is:
    [tex]\vec{E}= \frac{1}{4 \pi \epsilon_0} \sum \frac{q_i}{{r_{i0}}^2} \hat{r_{i0}} [/tex]
    What you have done so far is correct. I think first you should try the simple case of when there is just one charge. Also, since the system has translational symmetry, you can (without loss of generality), say that the position of the single charge is at the origin. So this will simplify your equation nicely.

    Once you've done this, you have the electric field (from Coulomb's law). So now, you need to use that in Gauss' law, and show that Gauss' law is satisfied. Or, equivalently, you can use it in the integral form of Gauss' law, and show that it is satisfied for any arbitrary region.

    Hint: the proof is mostly just maths. another hint: you will need to make use of the properties of the Dirac-delta function.

    Edit: It is not difficult to extend to the case when there are several stationary point charges. But (at least for me, anyway), it is best to think about the simple case first.

    Another Edit: Also, if you have not learned about Dirac-delta functions yet, you can still show that the integral form of Gauss' law holds for a spherical Gaussian surface centred on the point charge. But of course, this is not 'proving' Gauss' law fully.
     
    Last edited: Jan 7, 2013
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