Charge density of some potentials

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SUMMARY

The discussion focuses on the calculation of charge density from the potential V = exp(-λr)/r, where λ is a constant. It highlights the necessity of using the Laplacian operator, ∇²V, to derive charge density without encountering Dirac's delta function. The conversation emphasizes that for the point charge potential V = 1/r, the Laplacian results in ∇²V = 0 for r > 0, but leads to ∇²(1/r) = δ(r) when considering distributions. Understanding when to apply each method is crucial for accurate calculations in electrostatics.

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  • Understanding of electrostatics and potential theory
  • Familiarity with the Laplacian operator and its applications
  • Knowledge of Dirac's delta function and theory of distributions
  • Basic calculus, particularly partial derivatives
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hokhani
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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?
 
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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.
 
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