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Charge density of some potentials

  1. Jan 6, 2014 #1
    Suppose that we have the potential [itex]V=\frac{exp(-\lambda r)}{r}[/itex] that [itex]\lambda[/itex] is a constant. To calculate the charge density we have to calculate the [itex]\nabla^2V[/itex]. We can calculate directly by the formula [itex]\nabla^2 V =1/r^2 \frac{\partial (r^2 \frac{\partial V}{\partial r})}{\partial r}[/itex]without encountering the Dirac's delta function while If we calculate by another way we would have the Dirac's delta function in the final formula for the charge density. What we have to do not to encounter this discrepancy? Or how to understand which time we must use which formula?
    Last edited: Jan 6, 2014
  2. jcsd
  3. Jan 6, 2014 #2


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    Gold Member

    For the point charge potential V = 1/r, the same calculation will also lead to ∇²V = 0 .
    However, this Laplacian can only be calculated by derivation for r>0 .
    You need to go back to the theory of distributions.
    This will bring you to the result that ∇²(1/r) = δ(r) which cannot be obtained by elementary derivation.
    Last edited: Jan 6, 2014
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