Charge density of some potentials

AI Thread Summary
The discussion focuses on calculating the charge density from the potential V = exp(-λr)/r by using the Laplacian operator ∇²V. It highlights that calculating ∇²V directly avoids encountering the Dirac delta function, while an alternative method introduces it into the final formula. The conversation emphasizes the need to understand when to use each calculation method, particularly in relation to point charge potentials like V = 1/r, which yields ∇²V = 0 for r > 0. The necessity of returning to the theory of distributions is noted, as this approach reveals that ∇²(1/r) = δ(r), a result not achievable through elementary derivation. Understanding these nuances is crucial for accurate charge density calculations in potential theory.
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Suppose that we have the potential V=\frac{exp(-\lambda r)}{r} that \lambda is a constant. To calculate the charge density we have to calculate the \nabla^2V. We can calculate directly by the formula \nabla^2 V =1/r^2 \frac{\partial (r^2 \frac{\partial V}{\partial r})}{\partial r}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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