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Proof that all charge lies at the surface of a conductor

  1. Jun 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Use Gauss' Law to show that all static charge on a conductor lies entirely on the surface of the conductor.



    2. Relevant equations

    at a point outside the shell, this would be the final formula supposedly. E(r)= [itex]\frac{Q_s}{4\pi r^2 \epsilon}[/itex]

    3. The attempt at a solution

    Sorry if this is in the wrong section, but its technically a homework question.

    The proof of this goes like:

    From Gauss' law of Gravitation, we know all gravitational force acts on the outside of a spherical shell, because all the mass lies on the outside of the shell.

    If we apply Gauss' law of gravitation to a spherical shell, we get:

    [itex]\Gamma(4\pi r^2)[/itex] = Ms
    [itex]\rightarrow[/itex] [itex]\Gamma (r) [/itex] = -[itex]\frac{M_s}{4\pi r^2}[/itex]

    Thus, using the gravitational flux density formula ( g(r) = 4[itex] \pi[/itex]r2 )
    We find that the gravitational field strength at point P is :
    g(r)= -[itex]\frac{G M_s}{ r^2} [/itex][itex]\widehat{r}[/itex]

    Since all mass is contained on the outside, if we apply Gauss' law to any point on the inside of the shell, we get a field strength of 0, as so.

    [itex]\Gamma (4\pi r^2) [/itex] = 0
    [itex]\Gamma (r) [/itex] = 0
    [itex]\rightarrow[/itex] g(r) =0

    If we are to apply the following transformations to the above, this can be written in terms of electric charge.:

    g(r)[itex]\Rightarrow[/itex] E(r)
    G[itex]\Rightarrow[/itex] [itex]\frac{1}{4\pi \epsilon}[/itex]
    Ms [itex]\Rightarrow[/itex] Qs

    This gives us:
    E(r)[itex]\frac{Q_s}{4\pi \epsilon r^2}[/itex] for a point on the surface of the shell
    and
    E(r)= 0 for a point inside the shell.

    Apparently that is the solution, but it seems very.... lazy I guess :confused: Is that enough info on it? Or am i missing a big problem? Or can I sign Q.E.D and be done with it? :tongue:
     
    Last edited: Jun 26, 2013
  2. jcsd
  3. Jun 26, 2013 #2

    collinsmark

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    My comments are:
    • Why start with Gauss' Law for gravitation? Why not simply start (and end) with the regular Gauss' Law which is already all set up for electrostatics?
    • Your method, as you've done it, works for finding the charge inside a hollow, spherical shell. But can you use Gauss' Law to show that the static charge inside any conducting material is zero, regardless of its configuration? (Be it a cube, a cylinder, oblique shape, etc.)

    Gauss' Law is

    [tex] \oint_S \vec E \cdot \vec{dA} = \frac{Q_{\mathrm{enc}}}{\varepsilon_0} [/tex]

    The left hand side of the equation is a closed surface integral. To understand the equation, first, you'll need to imagine some sort of closed surface, whatever shape it might be. It can be of any shape, as long as it's a closed surface. If you take the dot product of an infinitesimal segment of area with the electric field at that location, and then integrate all those individual dot products over all the area in the closed surface, the result is the right hand side of the equation: The charge enclosed within the surface, divided by [itex] \varepsilon_0 [/itex].

    If you know* that the electric field within a solid, conducting material is always zero (and assuming that the material doesn't have any hollow cavities with charge inside), what does that say about the enclosed charge within the solid, conducting material?

    *(Hint: you need to know beforehand that the electric field within a solid, conducting material is always zero for this to work.)
     
  4. Jun 27, 2013 #3
    Ah... thats why it seemed lazy :tongue: it only solves for one type of conductor :smile: it was the only proof our lecturer gave us, but I just felt it was incomplete... which it was in a sense.

    If there is a charge on the material, it has to be coming from somewhere, namely the surface... because no electric field exists inside the material.

    Since an electric field is caused by the presence of charged particles, you could also say that no charged particles exist inside-they all reside on the surface of the enclosed material.
     
  5. Jun 27, 2013 #4

    WannabeNewton

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    Yes. It is as simple as that. Put mathematically, ##E = 0## within the material of the conductor hence ##\int E\cdot dA = 0 = Q_{\text{enclosed}}##. There was no need to make it as complicated and specialized as in post #1 :)
     
  6. Jun 27, 2013 #5

    BruceW

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    yeah, that's right. And to be even more specific, why must there be no electric field inside a conductor?
     
  7. Jun 27, 2013 #6
    Love of god... Proof in, what 2 lines :tongue: Thank you :)

    @Bruce
    is it:
    Cause the charge particles cause the electric field and, since these charges are of like sign (+/-) They repel as far away as they can... that is, the surface of the conductor. which means all electric field is caused at the surface, not inside.

    OR

    Because if the electric field inside wasnt zero, the conductor would have moving charges. These moving charge readjust so the electric field is zero :smile:

    Seriously, thanks for the help though :smile:
     
  8. Jun 28, 2013 #7

    BruceW

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    yep. The second one. using the nice terminology, we assume that the system is in electrostatic equilibrium, therefore the electric field inside the material must be zero (otherwise the free charges will move). But, since the charges cannot leave the material, it is OK for there to be a nonzero electric field at the surface of the material, since the charges are not free to move in the direction perpendicular to the boundary. And consequently, it is possible to have nonzero charge density at the surface of the material, while the system is still in electrostatic equilibrium.
     
  9. Jun 28, 2013 #8
    Perfect :smile: the second one makes more sense when you say it that way. :smile:
    thanks to all three of you :biggrin:
     
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