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Laplacian of electrostatic potensial

  1. Feb 11, 2010 #1
    Something occured to me just now. A question about the scalar potential.

    First I will do some calculations of the laplacian of the scalar potential in different electrostatic situations to give myself a basis for my question.

    Point charge:
    [tex]\phi =\frac{1}{4\pi\epsilon_0} \frac{q}{r}[/tex]

    [tex]\nabla^2\phi=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \phi}{\partial r}\right)=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\left(-\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}\right)\right)=\frac{1}{r^2}\frac{\partial}{\partial r}\frac{q}{4\pi\epsilon_0}=0[/tex]

    Line of uniform charge density:
    [tex]\phi=\frac{1}{2\pi\epsilon_0}\lambda \ln r[/tex]

    [tex]\nabla^2\phi=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial \phi}{\partial r}\right)=\frac{1}{r}\frac{\partial}{\partial r}\left(r\left(\frac{1}{2\pi\epsilon_0}\lambda\frac{1}{r}\right)\right)=\frac{1}{r}\frac{\partial}{\partial r}\left(\frac{1}{2\pi\epsilon_0}\lambda\right)=0[/tex]

    Plane of uniform charge density:

    [tex]\nabla^2\phi=\frac{\partial^2\phi}{\partial z^2}=0[/tex]

    My question then arises: Is the laplacian of the scalar potential always zero?

    If no, please show a counterexample.

  2. jcsd
  3. Feb 11, 2010 #2
    If I understand your question correctly, then no, in general the Laplacian of the scalar potential is not 0. Remember that:

    [tex]\vec{E} = -\vec{\nabla} \phi[/tex]

    And that...

    [tex]\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex]


    [tex] \frac{\rho}{\epsilon_0} = \vec{\nabla} \cdot \vec{E} = \vec{\nabla} \cdot (-\vec{\nabla}\phi) = -\nabla^2 \phi[/tex]
    Last edited: Feb 11, 2010
  4. Feb 11, 2010 #3
    I realize this, but then how are my calculations above incorrect? I took the laplacian of the point charge potential. Shouldn't this then get me [tex]\frac{3q}{4\pi\epsilon_0r^3}[/tex] ?
  5. Feb 11, 2010 #4
    So what gives with your examples? In short, I believe it is because you are dealing with charge distributions that involve singularities. The Laplacian doesn't behave well because of this.

    Take the analogous case of the divergence of the E-field of a point charge:

    [tex]\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}[/tex]

    On one hand, we know that the divergence of the point charge isn't zero. But on the other, when we actually compute the divergence:

    [tex]\vec{\nabla}\cdot\vec{E} = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \right) = \frac{1}{r^2}\frac{\partial}{\partial r}\left( \frac{q}{4\pi \epsilon_0} \right) = 0[/tex]

    So it appears something catastrophic has happened. But the trick is to realize that the magnitude of the electric field is infinite at the origin. To get the actual divergence, we must use the Divergence Theorem.

    See this page for more: https://www.physicsforums.com/showthread.php?t=165292"
    Last edited by a moderator: Apr 24, 2017
  6. Feb 11, 2010 #5
    Thank you, that thread cleared things up.
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