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Gauss's law and divergence

  1. Sep 21, 2014 #1
    Has anyone read the book by Daniel Fleisch, 'A Student's Guide to Maxwell's Equations'? I'm having some trouble with Chapter 1, page 36.

    He's talking about the divergence of an electric field originating from a point charge. Apparently, the divergence of the vector electric field is zero, because the spreading out of the field lines (as they get further away from the origin) is compensated for by the 1/R^2 reduction in the amplitude of the field. I don't really understand this?

    When I picture it, the field lines spread out so it must be diverging? How does the decreased amplitude help here?
    Last edited: Sep 21, 2014
  2. jcsd
  3. Sep 21, 2014 #2
    Also, if someone could tell me how to make LaTeX code work in these posts, I'd love you forever :rolleyes:
  4. Sep 21, 2014 #3
    First of all, the physical significance of divergence is not just spreading out. Refer the following pictures:-
    The above picture has 2 parts.
    In (b) part, the vectors are not spreading out from a point but as they approach more their magnitude increases (length of the arrow). Divergence is not just spreading, a vector space is said to be divergent if the vectors on average have an increase or decrease in magnitude along a particular direction. In (a) part, the vectors have 0 divergence thus.
    for the second pic,
    Interpret based on the above principle.

    Now, coming back to electric field, it is an inverse square field, by plotting we find that the vectors spread out, but notice that their magnitude too decreases proportionally with increasing radial distance, that's why, there is 0 divergence.

    divergence is non-zero when magnitude of vectors increase/decrease without proportion to decreasing/increasing distance respectively. Otherwise, it is 0.
    Divergence is far more from just spreading.
  5. Sep 21, 2014 #4


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    It's a misleading name. Divergence at some point basically tells you if there are field lines stating/ending at that point.

    Think in terms of sink/source of flux. If you consider a infinitesimal volume around a point, then divergence at that point is the net flux into/out of that volume. For the electric field near a point charge this is zero. Only at the point charge position there is a non-zero flux into/out of the infinitesimal volume around it.

    It means that the density of the field lines is higher on one side of the infinitesimal volume, but the amount of field lines going in still equals the amount going out. They are just spread over different areas.
    Last edited: Sep 21, 2014
  6. Sep 22, 2014 #5
    So, even though for a radial field the vectors are spreading out spatially (getting wider and wider apart), because they are also decreasing in magnitude this is effectively 'cancelled out' and so the divergence is zero? Because the overall vector field through an area further away from the origin will be the same because it's spread over a larger area?
  7. Sep 22, 2014 #6
    The net flux around a point charge is zero? I thought it was nonzero (depending on the sign of the charge, positive or negative flux)?
  8. Sep 22, 2014 #7


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    Not "around a point charge", but "around a point near a point charge". If your infinitesimal volume includes the point charge, the net flux it's not zero. But for any point that isn't exactly the point charge position, a infinitesimal volume around it doesn't contain any charge, so the incoming flux equals the outgoing flux.
  9. Sep 22, 2014 #8
    Ohhhhh. Right, got it. That makes more sense!
  10. Sep 22, 2014 #9
    Let me see if I've got this straight in my brain:

    Flux is the 'amount' of the vector field flowing though a surface in unit time. (Rate of flow.)
    Divergence is the net flux through a unit volume.

    The reason that a radial vector field with [itex]\begin{equation} \frac{1}{r^2} \end{equation}[/itex] amplitude reduction is zero is because the net flux per unit volume is not changing. Although the field lines are decreasing in amplitude, the summing over the larger volume compensates for this reduction and so the divergence is zero.

    Is that right?
  11. Sep 22, 2014 #10


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    First of all, one should define all quantities discussed properly. The flux of a vector field through a closed surface, [itex]\partial V[/itex], which is the boundary of a volume in space, [itex]V[/itex] is given by the surface integral
    [tex]\Phi(\vec{E},\partial V)=\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}.[/tex]
    By convention, the surface-normal vectors [itex]\mathrm{d}^2 \vec{f}[/itex] are always pointing outwards, way from the volume [itex]V[/itex].

    The divergence of a vector field at a point [itex]\vec{r}_0[/itex] is given by the limit
    [tex]\vec{\nabla} \cdot \vec{E}(\vec{r}_0)=\lim_{V(\vec{r}_0) \rightarrow \{\vec{r}_0 \}} \frac{1}{\Delta V(\vec{r}_0)} \int_{\partial V(\vec{r}_0)} \mathrm{d}^2 \vec{F} \cdot \vec{E}.[/tex]
    In some sense you can say that the divergence is the local form of the flux per unit volume through a closed surface around the point in question.

    For the Coulomb field,
    [tex]\vec{E}=\frac{q}{4 \pi r^2} \frac{\vec{r}}{r}[/tex]
    you can easily figure out that for any [itex]\vec{r} \neq \vec{0}[/itex]
    [tex]\vec{\nabla} \cdot \vec{E}(\vec{r})=0[/tex]
    but that the divergence at the origin (i.e., the place where the point charge sits) it's diverging to infinity.

    What you also can easily calculate (at least for spheres around the origin) is that
    [tex]\Phi(\vec{E},\partial V)=q,[/tex]
    for any volume containing the origin, while this flux vanishes for any volume that does not contain the origin.

    For an arbitrary charge distribution this leads to Gauss's Law, which reads in integral form
    [tex]\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 \vec{r} \rho[/tex]
    and in local form, by just using the definition of the divergence given above
    [tex]\vec{\nabla} \cdot \vec{E}=\rho.[/tex]
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