The Divergence of a Regularized Point Charge Electric Field

[1missing]

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!

 

[member 587159]

What you wrote is completely correct. At first sight, calculating this looks like a lot of work, but once you realize how symmetrical the expression you wrote down is, you can easily conclude that it suffices to calculate one of the three derivatives.

 

[1missing]

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?