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Divergence of flux density

  1. Jul 23, 2009 #1
    Question:
    Can someone remind me why the divergence of the electric flux is equal to the volume charge density,
    [tex]\nabla \bullet \vec{D} = \rho_{v}[/tex] (where [tex]\vec{D}[/tex] is the electric flux density).

    Thoughts:
    The divergence measures the flow of a field out of a region of space. The del operator takes the gradient of the field, which measures the tendency of the field to diverge away in space (or the opposite). So when we take the divergence of the electric flux density, we are measuring how quickly the tendency of the flux to diverge in a given space. But how is that the volume charge density? Isn't charge density entirely different from the divergence of the electric flux?
    Thanks,


    JL
     
  2. jcsd
  3. Jul 23, 2009 #2
    Isn't that the differential form of Gauss's Law with free charge? You could get to that starting with the integral form of Gauss's Law and using the divergence theorem. You can't really prove Gauss's law, or at least I didn't think you could.
     
  4. Jul 23, 2009 #3

    jgens

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    Gold Member

  5. Jul 23, 2009 #4
    Here's how you go from the integral form of Gauss's law for free charge to the differential form.

    [tex]\oint \textbf{D} \cdot d\textbf{A}=Q_{f}(V)[/tex]

    By the divergence theorem:

    [tex]\oint \textbf{D} \cdot d\textbf{A}=\int(\nabla \cdot \textbf{D})dV = Q_{f}(V)[/tex]

    Since the [tex]Q_{f}[/tex] is just net free charge enclosed in the Gaussian surface, you can say that it's just the integral of the volume charge density.

    [tex]Q_{f}(V)=\int\rho_{f}dV[/tex]

    [tex]\int(\nabla \cdot \textbf{D})dV = \int\rho_{f}dV[/tex]

    [tex]\nabla \cdot \texbf{D} = \rho_{f}[/tex]
     
  6. Jul 23, 2009 #5

    ideasrule

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    Homework Helper

    To the contrary, the integral form of Gauss's law is easy to prove with Coloumb's law, and Coloumb's law can easily be arrived at from intuition. To begin, prove that Gauss's law is true for a sphere centered on a single point charge. Electric field decreases as the square of the distance and the perpendicular projection of the area subtended by a certain solid angle increases as the square of the distance, so the product of the two remains constant. QED.
     
  7. Jul 23, 2009 #6
    Gauss's law in differential form is,
    [tex]\nabla \bullet \vec{D} = \frac{\rho_{v}}{\epsilon_{0}}[/tex]

    We also know [tex]\epsilon = \epsilon_{r} \epsilon_{0}, \vec{D} = \epsilon \vec{E}[/tex] respectively.

    So it follows [tex]\epsilon[\nabla \bullet \vec{E}] = \nabla \bullet \vec{D} = \frac{\epsilon \rho_{v}}{\epsilon_{0}} = \epsilon_{r} \rho_{v}\neq \rho_{v} ?[/tex]

    Can someone help me with this?
     
    Last edited: Jul 23, 2009
  8. Jul 23, 2009 #7
    Oh yeah I forgot about that, lol. Lewin did it in his lecture.
     
  9. Jul 23, 2009 #8
    That's not the differential form of Gauss's law with respect to free charge. It is:

    [tex]\nabla \cdot \textbf{D} = \rho_{f}[/tex]
     
  10. Jul 23, 2009 #9
    I forgot to mention that [tex]\textbf{D}[/tex] alone is not the electric flux density. The vector [tex]\textbf{D}[/tex] represents the electric displacement field. The divergence operator gives you a scalar value that often is called the flux density. Hence, [tex]\nabla \cdot \textbf{D}[/tex] is called the electric flux density.
     
  11. Jul 23, 2009 #10

    According to my notes, and this web-page, D is the electric flux density,
    http://encyclopedia2.thefreedictionary.com/Electric+flux+density?
     
  12. Jul 23, 2009 #11
    Never mind, found my error, thanks for going along with the process.
     
  13. Jul 23, 2009 #12
    Anytime. :-)
     
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