Divergence in spherical polar coordinates


by Idoubt
Tags: coordinates, divergence, polar, spherical
vanhees71
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#19
Aug19-11, 06:26 AM
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Quote Quote by Idoubt View Post
This by definition is a singularity isn't it?
No, it's a distribution, the socalled Dirac [itex]\delta[/itex] distribution. It's a very useful concepts. It's even so useful that the mathematicians built a whole new subfield of mathematics to define this concept rigorously, functional analysis!
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#20
Aug19-11, 06:31 AM
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I suspect you're referring to Maxwell's law that says:
[tex]\textrm{div }\vec E = \frac {\rho} {\epsilon_0}[/tex]

The divergence of a spherically symmetric charge is zero everywhere except at the place where the charge actually is.
A point charge is a special case, since the charge density would be infinite (actually a Dirac delta-function).
In practice the charge would take in a certain volume.
The divergence of the electric field is not zero at the place where the charge density is not zero.
Idoubt
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#21
Aug19-11, 08:58 AM
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So can I go one more step and state that, at the points in space where the charge 'is' the electric field can no longer be defined by the function [itex]\frac{1}{r^2}[/itex][itex]\hat{r}[/itex] ?
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#22
Aug19-11, 09:44 AM
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Correct.
In the case of a solid sphere with constant charge density within, the electric field is proportional to [itex]r \hat {\boldsymbol r}[/itex] (inside the sphere).
Idoubt
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Aug19-11, 10:08 AM
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thank you, this helped a lot


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