Gauss' Law and Coulomb's Law — Question 6

Question 6

A modified inverse-power field would require distributed charge outside its source

Consider the hypothetical radial field $E_r(r)=kq r^{-2}(r_0/r)^\delta$ for $r>0$, with $q,r_0>0$ and constant $\delta$. Keep the ordinary Gauss law and $k=1/(4\pi\epsilon_0)$.

1. Find the effective enclosed charge at radius R and the net charge between radii a and b, where $0<a<b$.

2. Derive the ordinary charge density for $r>0$ from $dQ_{\mathrm{enc}}=4\pi r^2\rho\,dr$. Determine its sign for $\delta>0$ and identify the exponent that allows a charge-free exterior.

Question 6 - Solution

A modified inverse-power field would require distributed charge outside its source

(a) Method and calculation

$$\Phi(R)=4\pi R^2E_r(R)=\frac q{\epsilon_0}(r_0/R)^\delta.$$$$Q_{a<r<b}=q\left[(r_0/b)^\delta-(r_0/a)^\delta\right].$$

(b) Interpretation and further calculation

$$\rho(r)=\frac1{4\pi r^2}\frac{d}{dr}\left[q(r_0/r)^\delta\right]=-\frac{\delta q r_0^\delta}{4\pi r^{3+\delta}}.$$$$\delta>0\Rightarrow\rho<0;\qquad\rho=0\text{ for all }r>0\Longleftrightarrow\delta=0.$$Under ordinary Gauss law, changing the inverse-square exponent changes the implied source distribution.

Physical check

This is a consistency test for a hypothetical field, not a proposed modification of measured electrostatics.