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Mean Value Theorem of Electrostatics

  1. Jan 12, 2006 #1
    In "Classical Electrodynamics - 3rd Ed.," J.D. Jackson has an exercise, 1.10, to derive the mean value theorem of electrostatics. Does anyone know of a derivation which is located on the web?

    Pete
     
  2. jcsd
  3. Jan 12, 2006 #2

    dextercioby

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    Yes, this one

    "The average value of the potential over the spherical surface is

    [tex] \bar{\Phi} =\frac{1}{4\pi R}\int \Phi \ dA [/tex]

    If u imagine the surface of the sphere as discretized, u can rewrite the integral as an infinite sum

    [tex] \frac{1}{a}\int \ dA =\Sigma_{area} [/tex]

    Then, takin the derivative wrt to R

    [tex]\frac{d\bar{\Phi}}{dR}=\frac{d}{dR}\Sigma \Phi=\Sigma \frac{d\Phi}{dR} [/tex]

    Convert the infinite sum back to an integral & get

    [tex] \frac{d\bar{\Phi}}{dR}=\frac{1}{4\pi R^{2}} \int \frac{d\Phi}{dR} \ dA [/tex]

    Now, using that

    [tex] \frac{d\Phi}{dR}=-E [/tex] and Gauss's law (no charge inside the sphere) that gives

    [tex] \frac{d\bar{Phi}}{dR}=0 \Rightarrow \bar{\Phi}_{surface}=\Phi_{center} [/tex]

    Q.E.D.


    Daniel.
     
  4. Jan 14, 2006 #3
    No. It isn't. Its
    [tex] \bar{\Phi} =\frac{1}{A}\int d\Phi [/tex]
    where A is the area of the sphere.

    Pete
     
  5. Jan 14, 2006 #4
    This isn't quite right either.

    It isn't clear what you mean by [itex] \int d\Phi [/itex]
    since usually the integral of an exact differential
    is just the function evaluated at the endpoints.

    Or alternatively
    [tex] \int d\Phi = \int \frac{\partial \Phi}{\partial x} dx +
    \int \frac{\partial \Phi}{\partial y} dy +
    \int \frac{\partial \Phi}{\partial z} dz [/tex]
    which is, of course, a line integral (which isn't what we
    want).


    This doesn't have the right dimensions. On the left
    we have dimensions of [Phi]; and on the right we have
    dimensions of [Phi]x[Area]/[Length].

    I propose another formula:
    [tex]
    \bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA
    [/tex]

    This also has the advantage that if Phi is constant
    on the sphere we have
    [tex]
    \bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA =
    \frac{1}{4 \pi R^2} \Phi \int_A dA =
    \frac{1}{4 \pi R^2} \Phi ( 4 \pi R^2 ) =
    \Phi
    [/tex]
     
  6. Jan 15, 2006 #5
    That would be true if you were talking about a line integral rather than in this case where the integral is a surface integral. What [itex]d\Phi[/itex] is? Hmmm. Perhaps you're right. I can't say that is a small element of [itex]\Phi[/itex] since that makes no sense to me at the moment. Thanks.

    Pete
     
  7. Jan 15, 2006 #6

    robphy

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    May I propose a more [notationally] precise formula:
    [tex]
    \bar{\Phi} = \frac{1}{A} \int_A \Phi dA
    [/tex]
    or a more [conceptually] precise formula for this average:
    [tex]
    \bar{\Phi} = \frac{ \int_A \Phi dA } {\int_A dA}
    [/tex]
     
  8. Jan 16, 2006 #7
    Ah, pedantry, thou most noble virtue. You are, of course, correct. Now the next person (who reads this thread) who needs a 2-D average will know the correct generalization.
     
    Last edited: Jan 16, 2006
  9. Jan 16, 2006 #8

    dextercioby

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    I admit, it was R^2 in the denominator...

    Daniel.
     
  10. Jan 16, 2006 #9

    robphy

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    This is a Homework subforum... and, in my experience, my students appreciate seeing the general idea in a consistent memorable notation rather than a special case. After all, there was some confusion with this formula.

    Although the notation "A" [implicitly] suggests an area average, it is certainly not necessarily so.
     
  11. Jan 16, 2006 #10
    Hi Rob

    Do you know where I can find a solution? Jackson seems to want the student to use Green's theorm to solve this.

    Pete
     
  12. Jan 7, 2010 #11
  13. Jan 7, 2010 #12
    You do realize that this post is coming onto 4 years old? That the original poster hasn't asked a question for well over a year?

    There is usually no reason to resurrect threads that are more than 2 months old, let alone ones that are 4 years old.
     
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