Divergence of a vector field

  • Thread starter 1missing
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  • #1
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1. Problem: Consider vector field A##\left( \vec r \right) = \frac {\vec n} {(r^2+a^2)}## representing the electric field of a point charge, however, regularized by adding a in the denominator. Here ##\vec n = \frac {\vec r} r##. Calculate the divergence of this vector field. Show that in the limit a -> 0 the divergence becomes proportional to the δ-function.

Homework Equations


∇⋅ = ## \frac \partial {\partial x} + \frac \partial {\partial y} + \frac \partial {\partial z}##

The Attempt at a Solution


So it seemed pretty straight forward to me, but I feel like there's something fundamental that I'm not seeing.

##\vec r = \left( x, y, z\right)##

##r = \sqrt {x^2 + y^2 + z^2}##

∇⋅A## \left( \vec r \right) = {\frac \partial {\partial x}} \frac x {\left( x^2 + y^2 + z^2\right)^{1/2} \left( x^2 + y^2 + z^2 + a^2 \right)} + {\frac \partial {\partial y}} \frac y {\left( x^2 + y^2 + z^2\right)^{1/2} \left( x^2 + y^2 + z^2 + a^2 \right)} + {\frac \partial {\partial z}} \frac z {\left( x^2 + y^2 + z^2\right)^{1/2} \left( x^2 + y^2 + z^2 + a^2 \right)} ##

I don't have any trouble with the computation, rather I feel like I didn't set this up correctly. Can anyone confirm if I'm moving in the right direction? Thanks!
 

Answers and Replies

  • #2
member 587159
What you wrote is completely correct. At first sight, calculating this looks like a lot of work, but once you realise how symmetrical the expression you wrote down is, you can easily conclude that it suffices to calculate one of the three derivatives.
 
  • #3
12
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Awesome, guess I'll start plodding away. And I guess just to make sure, for the second part, all I would need to do is set a = 0, integrate over all space, and confirm it equals 1?
 

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