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