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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?

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Here's Coulomb's Law:

http://upload.wikimedia.org/math/7/a/6/7a655cf7d117c188b73bec1c82077218.png

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.

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This is a valid proof, but I doubt that it is the scale invariance symmetry meant by Prof. Preskill. He wrote: "Actually, this problem can also be solved by a symmetry argument, though the symmetry used is less obvious than rotational invariance. Readers may enjoy constructing this symmetry argument, which (in keeping with the theme of this post) I find more elegant than the argument using concentric rings suggested by your hint." Please read the discussion in the post in the question. –  Tarek Oct 5 '12 at 5:53
    
I see. I think josh wrote the answer you want. –  Mark Eichenlaub Oct 5 '12 at 12:52
    
I edited this answer to reiterate the argument in a slightly-different way. –  Mark Eichenlaub Oct 5 '12 at 14:04
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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.

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Great answer! If you have a volume distribution of charge and you scale everything up by $\lambda$, the electric field scales as $\lambda$ as well. With a 2-d charge distribution we have this scale invariance property. –  Mark Eichenlaub Oct 5 '12 at 12:57
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