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Gauss again

  1. Feb 1, 2005 #1

    quasar987

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    I asked the following question to my teacher through email a week ago, and he hasn't replied.

    Consider a uniformly charged spherical shell of radius R and areal density [itex]\sigma[/itex] and consider also a cartesian system whose origin coincide with the center of the shell.

    Clearly if we attempt to calculate the field at r = R directly through Coulomb's law, we obtain that the field is undefined, because we'd have a (r' - r') at the denominator (equ. 2.7 in Griffiths)

    But, as Griffiths remarks (pp.88), if we apply Gauss' Law, we obtain that the field at r = R is of magnitude [itex]\sigma \epsilon_0[/itex]. So what should we think? Is the sclalar function E(r) undefined at r = R or not ?!
     
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  3. Feb 1, 2005 #2

    dextercioby

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    It is defined....Naturally,since it has a finite value...Compute the field inside the sphere and outside and then plot the graph...Do you see anything interesting??

    Daniel.

    P.S.How do you compute the field using Coulomb's law?
     
  4. Feb 1, 2005 #3

    quasar987

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    Well it's not so simple I think, since, like I said, using coulomb's law, it DO comes out undefined. Both assistant teachers said it was undefined, but I don't trust them.

    Griffith's equ. 2.7 I have written in post #13 of the infamous Gauss' Law proof thread. I would copy it but for some reason, clicking the formula does nothing so I can't retrieve the code.

    You can see, in the integral, if [itex]\vec{r} = \vec{r'}[/itex], it is undefined.
     
  5. Feb 1, 2005 #4

    dextercioby

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    Are u referring to poste #6 and its first formula???
    If so,then I have 2 objections...
    a)That is not Coulomb's law...
    b)You cannot use that formula...

    Which formula need you use?

    Daniel.
     
  6. Feb 1, 2005 #5

    quasar987

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    Yes, I'm referring to the scalar equation of the field from post #6. That is the expression Griffiths gives for the field produced by a continuous distribution of charge.
     
  7. Feb 1, 2005 #6

    dextercioby

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    Yes,that's true.But that one is valid for [itex] \rho (\vec{r'}) [/itex] which is A VOLUME DENSITY OF CHARGE.You need another formula for a SURFACE DENSITY OF CHARGE...

    Daniel.
     
  8. Feb 1, 2005 #7

    quasar987

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    Oh. Right. Well. This one is given right before. It has exactly the same form, except

    1) The integral is over a surface.
    2) [itex]\rho(\vec{r'})[/itex] is replaced by [itex]\sigma(\vec{r'})[/itex]
    3) dV is replaced by dA
     
  9. Feb 1, 2005 #8

    krab

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    It is undefined. It is related to the fact that an infinitely thin shell is unphysical. In the limit, the field has a discontinuity at r=R, but in real life, the shell does have a thickness and the field varies smoothly from zero inside the shell to its max value just outside.
     
  10. Feb 1, 2005 #9

    quasar987

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    Mmh, but how should one interpret the result provided by Gauss' Law:
     
  11. Feb 2, 2005 #10

    Galileo

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    At theboundary of a charged surface there's always a discontinuity in the electric field. Specifically in the normal component of the electric field (normal to the surface). The parallel component is continuous.

    I think the relation was:

    [tex](\vec E_1 - \vec E_2)\cdot \vec n = 4\pi \sigma[/tex]
    or
    [tex](\vec E_1 - \vec E_2)\cdot \vec n = \sigma/\epsilon_0[/tex]
    depending on your choice of units.
    The n is the unit normal to the surface.

    Therefore strictly the field is not defined. You always have a singularity in E precisely at the point where charge is.
    Since this is an approximation, as krab pointed out, in reality E is always smooth.
     
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