# Divergence of a vector field

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

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.

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?