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How to calculate the charge density in an electric field

  1. Jun 14, 2013 #1
    I've been studying electric fields in class for some time and one thing is seemingly contradictory and really confuses me.

    The charge density ρ is related to the electric field E and the permiativity ε and the potential [itex]\Phi[/itex] by the following equation

    ρ/ε=∇[itex]\cdot[/itex]E=-∇2[itex]\Phi[/itex]

    if we examine the electric field created by a single point charge of magnitude q located at the origin, then the electrostatic potential can be expressed as follows

    [itex]\Phi[/itex]=[itex]\frac{q}{4πε(x^2+y^2+z^2)}[/itex]

    now I would expect the charge density in this system to be zero everywhere except the origin but if we take the laplacean of this electric field, instead we get

    ρ/ε=-∇2[itex]\Phi[/itex]=[itex]\frac{2q}{4πε(x^2+y^2+z^2)^2}[/itex]

    which is clearly non-zero.

    is there an explanation for this discrepancy? Have I violated some fundamental assumption?

    thanks in advance
     
    Last edited: Jun 14, 2013
  2. jcsd
  3. Jun 14, 2013 #2

    WannabeNewton

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    Science Advisor

    Well you didn't really post your work so it's hard to tell where you went wrong. The Coulomb potential is much more elegantly expressed (and easier to work with) in spherical coordinates. ##\varphi = \frac{Q}{4\pi \epsilon_0 }\frac{1}{r}## hence ## \nabla^{2}\varphi = \frac{Q}{4\pi\epsilon_{0}}\frac{1}{r^{2}}\partial_{r}(r^{2}\partial_{r}\frac{1}{r}) = 0## for ##r \neq 0##.
     
  4. Jun 14, 2013 #3
    The potential goes like 1/r and not 1/r^2.
    Try the correct form of the potential. It works even in Cartesian coordinates even though is more work.
     
  5. Jun 14, 2013 #4
    thanks

    it seems like i was using the wrong equation for electric potential. i had an r2 in the denominator instead of just r
     
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