# Intuition for the charge density of a homogeneously charged disc

The charge distribution $$\rho(\vec{r})=\frac{q}{\pi a^2r}\theta{(a-r)}\delta{(\vartheta-\pi/2)}$$ describes a homogeneously charged disc which makes sense in terms of distributions. But if we plot the charge distribution it would decrease with greater radius. I know that the $$r$$ should only compensate a part of the Jacobian but the charge distribution makes intuitive no sense.

## 1 Answer

It can help to think of the delta-function as a limit of a "pulse" function with a fixed area under the curve. Suppose that instead of that delta-function in $$\theta$$, we replaced it with a fixed-integral pulse of width $$\epsilon$$ around $$\pi/2$$: $$P_\epsilon(\theta) = \begin{cases} 1/\epsilon & |\theta - \pi/2| < \epsilon/2 \\ 0 & \text{otherwise} \end{cases}$$ In the limit as $$\epsilon \to 0$$, this becomes $$\delta(\theta - \pi/2)$$; this is one of the definitions of the $$\delta$$-function.

But let's now think about the support of $$P_\epsilon(\theta)$$ in 3D space (the part of space where it's non-zero.) This would be a "wedge of space" including the equator, narrower near the origin and getting wider as $$r$$ gets bigger. In fact, the width of this wedge is proportional to $$r$$. So if we wanted to renormalize this function such that its integral over a sphere of constant $$r$$ was independent of $$r$$, we would have to use $$P_\epsilon(\theta)/r$$ instead of $$P_\epsilon(\theta)$$. In the limit as $$\epsilon \to 0$$, this would become $$\delta(\theta - \pi/2)/r$$ instead of $$\delta(\theta - \pi/2)$$.