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Electric potential for an infinite plane charge distribution

  1. Oct 5, 2009 #1
    Hello everybody,
    I have to calculate the electric field and the potential for a charge [tex]q[/tex] placed at distance [tex]d[/tex] from an infinite plane charge distribution [tex]\sigma[/tex].

    For the electric field there's no problem, but how I can get the electric potential for an infinite charge distribution?
     
  2. jcsd
  3. Oct 5, 2009 #2

    Vanadium 50

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    Integrate the field.
     
  4. Oct 5, 2009 #3
    ok, but I get an unknown constant.

    The field is [tex]E=\frac{\sigma}{2\epsilon_{0}}[/tex] in the x direction. So if I integrate it I get

    [tex]V(x)=V(0)-\frac{\sigma}{2\epsilon_{0}}x[/tex]

    Where the constant is unknown...maybe I don't need to know it

    Further, if I calculate the potential starting from a circular plane distribution and then pulling the radius to infinity I get an infinite potential
     
    Last edited: Oct 5, 2009
  5. Oct 5, 2009 #4

    gabbagabbahey

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    No, that's just the field of a uniformly charged plate with surface charge density [itex]\sigma[/itex].... what about the field of the point charge?

    There is always an "unknown constant" when calculating the potential. This is because it is determined from the differential equation [itex]\textbf{E}=-\mathbf{\nabla}V[/itex], and as you should know, first order DE's need at least one boundary/initial condition to find a unique solution...You are free to choose any value for your constant, simply by choosing a suitable reference point (a point where you define the potential to be zero)...in this case, choosing the origin as a reference point makes things simple (so that V(0)=0).


    Unless you show your calculation, I cannot be certain of your error, but I suspect you are unknowingly choosing your reference point to be at r=infinity...r=infinity is usually a bad choice of reference point when dealing with charge distributions the extend to infinity.
     
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