Scale invariance symmetry as a simple argument in an electrostatics problem

Question

In the comments to this post, it was hinted that proving that the force acting on a charge at a vertical distance from a uniformly charged plane is independent of that distance can be done by recalling the scale invariance symmetry. Can someone explain that to me?

Answer 1

Here's Coulomb's Law:

If you scale everything by $\lambda$, you get $\frac{1}{\lambda^2}$ in the denominator, but you must also introduce a Jacobian for the integral.

For a volume charge, the Jacobian is $\lambda^3$, so you're on the surface of a ball of charge and you make the ball bigger (with the same charge density), the E-field increases proportionately.

For a surface charge, though, the Jacobian is $\lambda^2$, which cancels the $\frac{1}{\lambda^2}$ in the denominator of Coulomb's law. Thus, a 2-d distribution of charge is "scale invariant". Since scaling an infinite sheet leaves it unchanged, scaling changes only the distance from the sheet, and we see that the E-field is independent of distance.

Answer 2

One way to see this is to just work the math out. Put a test charge Q at a vertical distance $z$ above a surface with a constant surface charge density $\sigma$. The potential clearly depends only on the vertical distance too: $$ \phi(z) = Q\sigma\int_0^{2\pi}\!d\theta \int_0^\infty\!dr\,\frac{r}{\sqrt{r^2+z^2}}\,. $$ Here $r$ is the radial coordinate in a polar coordinate system on the charged plane.

The electric field acting on the test charge is also only a function of the vertical distance (and also only has a component in the vertical direction): $$ E_z(z) = -\phi^\prime(z) = 2\pi Q\sigma \int_0^\infty\!dr\,\frac{rz}{(r^2+z^2)^{3/2}}\,. $$ Now, change the vertical distance $z \rightarrow \lambda z$, for some arbitrary (as long as it's not zero) scaling factor $\lambda$. Then, change variables of integration also, $r \rightarrow \lambda r$. \begin{align} E_z(\lambda z) &= 2\pi Q\sigma\int_0^\infty\!dr\,\frac{r\,(\lambda z)}{(r^2+(\lambda z)^2)^{3/2}}\\ &= 2\pi Q\sigma\int_0^\infty\!dr^\prime\,\frac{r^\prime z}{(r^\prime{}^2+z^2)^{3/2}}\\ &= E_z(z)\,. \end{align} The factors of $\lambda$ cancel out, i.e. there is a scaling invariance because a rescaling is seen to be irrelevant. As pointed out by Mark Eichenlaub, this hinged on the exact power-law behavior in Gauss' Law: $E \propto |x|^{-2}$ around a point source.