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Chargedistribution from a given potential

  1. Aug 10, 2009 #1
    The following potential is given
    10gb32x.png

    The question is what the charge distribution is. The middle part is a charged dielectric. The two discontinuous points are the result of a charge accumulated in one point. And after that point the potential doesn't vary. So my thoughts are that the physical situation is a charged dielectric between two charged plated, with the charges of the dielectric oposite to the charge on the plate it faces.

    I think I'm right so far. But now I want to calculate the charge distribution. The hint was to use delta-function and I can see why, but I don't know how. Can any of you help me?

    PS: My paint skills suck, but I hope it's clear that the middle parabolic and the left potential is higher than the right one.
     
  2. jcsd
  3. Aug 10, 2009 #2
    Use the equation ∆φ ~ ρ. In a 1D case the second derivative of your potential will give the charge density.
     
  4. Aug 10, 2009 #3
    But what is the equation for such a potential?
     
  5. Aug 10, 2009 #4
    Sorry, I should have written it as ρ ~ ∆φ (Gauss law) or ρ(x) ~ (d²/dx²)φ(x) in your case.

    ρ is a charge density and φ is the electrostatic potential. Depending on units, the equation may contain 4π, etc.
     
  6. Aug 10, 2009 #5
    I know how to solve a laplacian, but I can't find the equation for the potential.
     
  7. Aug 10, 2009 #6
    The equation is the following: ρ(x) ~ (d²/dx²)φ(x) in your case. All you have to do is to differentiate twice your potential given in your figure.
     
  8. Aug 10, 2009 #7
    I know all that. I've studied my book (Introduction to Electrodynamics), but I need the equation for the potential. That's my problem...
     
  9. Aug 10, 2009 #8
    You mean an analytical formula for your curve in the figure? Approximate it with something differentiable and you will obtain an approximate charge density.

    The differential equation for a potential is the Gauss law ∆φ ~ ρ.

    If the charge density ρ is given, you have to integrate this differential equation to find the potential φ.

    If the potential φ is given, you have to differentiate it to find the density ρ.
     
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