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My book has a section on symmetry operations.

It says, (if the plane of charge is the yz plane) translation symmetry along the y-axis and z-axis implies that the electric field is constant if one translates along the y and z axes respectively. Also, due to rotational symmetry, the field is is perpendicular to the yz plane. I understand this much.

Further, it says, another symmetry can be invoked to show that the field is independent of the x co-ordinate as well (without mentioning the symmetry).

I thought about translating the plane along the x-axis but it would change the charge distribution in space and hence, is not a symmetry operation. What is the symmetry the book mentions?

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3 Answers 3

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Scale symmetry. An infinite plane looks the same no matter how far away from it you are.

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    $\begingroup$ @GSmith Independence of field of $x$ only holds because of $1/r^2$ law. That isn't scale invariance. $\endgroup$
    – Elio Fabri
    Commented Feb 27, 2019 at 10:49
  • $\begingroup$ @ElioFabri The scale invariance I was referring to was the scale invariance of the geometry of an infinite plane. The standard argument goes as follows: The field of a infinite planar charge density $\sigma$ is obviously proportional to $\sigma$. In Gaussian units, $\sigma$ has the same units as electric field. Thus, if the field has any position dependence, it must depend on $x$ (the distance from the plane) divided by some length scale $a$: $(x/a)^{-p}$, $e^{-x/a}$, $f(x/a)$, etc. But for an inflnite plane, there is no length scale, so there can be no dependence on $x$. $\endgroup$
    – G. Smith
    Commented Feb 27, 2019 at 20:40
  • $\begingroup$ @GSmith IMO your argument relies on validity of Coulomb's $1/r^2$ law. For the sake of argument let's assume Coulomb's law were instead $F=q_1\,q_2/r^3$. Then the field of an infinite plane would be a function of $x$: $E=\pi\sigma/x$. No length scale is required. (Of course in these electrostatic units $\sigma$ and $E$ have different dimensions.) $\endgroup$
    – Elio Fabri
    Commented Feb 28, 2019 at 10:16
  • $\begingroup$ @ElioFabri Yes, my argument relies on the validity of Coulomb’s Law! But the point is you don’t have to do an integral over the plane. You can just use a dimensionality argument from Coulomb’s Law (charge per unit area has same dimensions as field) together with the scale invariance of the infinite plane. As the OP’s book said, “another symmetry can be invoked to show that the field is independent of the x co-ordinate.” $\endgroup$
    – G. Smith
    Commented Feb 28, 2019 at 17:08
  • $\begingroup$ @ElioFabri I think the problem is that I wrote an overly terse answer in haste. Thanks to your feedback, I think my comments have now made clear how to invoke Coulomb’s Law plus planar symmetry to conclude that the field must be constant in the $x$ direction. This argument can be found in various textbooks and I did not come up with it myself. $\endgroup$
    – G. Smith
    Commented Feb 28, 2019 at 17:11
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I'll try to elaborate on the previous answer and comments to show explicitly how the symmetry argument goes.

Let us start with the vanishing of the electric field's $y$ and $z$ components. It is a direct consequence of the infinite extension of the plane and the superposition principle for the electric forces. Indeed, the field at each space point of cartesian coordinates ($x,y,z$), due to a uniform planar charge in the $x=0$ plane, can be considered as the superposition of the fields originating from an infinite sequence of concentric circular crowns centered at $(y,z)$. The $y$ and $z$ components of the field at a $(x,y,z)$ must vanish because, for each element of the circular crown, there is an opposite element whose contribution to the components of the field parallel to the plane will cancel.

Notice that this argument requires only the rotational and translational invariance (in the plane) of the charge distribution and does not use the distance dependence of the Coulomb law.

On the contrary, we can prove the $x$-independence of the $x$-component of the field only for the Coulomb interaction, because only that form of interaction introduces a specific scale invariance. For the sake of simplicity, let's consider the contribution to the field at ($x,0,0$), due to the circular crowns centered at the origin of the $x=0$ plane. Due to the translational symmetry, the same argument applies to any other point ($x,y,z$). By using cylindrical coordinates ($r,\phi,x$) with the line $z=0$ and $y=0$ as cylinder axis, the contribution of the crown of width ${\mathrm d}r$ to the $x$-component of the field at ($x,0,0$) is $$ {\mathrm d}E_x= 2 \pi \frac{x r {\mathrm d}r}{(r^2+x^2)^{\frac32}}, $$ that is invariant under rescaling of $x$ and $r$ by the same factor: $$ {\mathrm d}E_x(\lambda x, \lambda r) = {\mathrm d}E_x(x,r). $$ Therefore, for each $x$, we can choose $\lambda=\frac{1}{x}$ to get $${\mathrm d}E_x(x,r)={\mathrm d}E_x(1,(r/x))=2 \pi \frac{ \left( \frac{r}{x} \right) {\mathrm d}\left( \frac{r}{x}\right) }{\left (\left( \frac{r}{x} \right)^2+1 \right)^{\frac32}}.$$ This implies that $ E_x(x)=\int_0^{\infty}{\mathrm d}E_x(x,r) $ is a constant. Any other interaction would imply an $x$-dependent $E_x$

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The potential of all points on a plane parallel to the sheet is same and the electric field vector at every point on such plane is same.

Now, because of the symmetry of the 3D space with respect to the infinite sheet, the field vector at $x=d$ is negative of that at $x=-d$ and is along positive and negative x-axes respectively.

Considering two cuboidal(or cylindrical, if you wish) Gaussian surfaces with two of the faces parallel to the sheet and at $x=d_1$ and $x=-d_1$ for one and at $x=d_2$ and $x=-d_1$, the charge enclosed in the surfaces is the same, so the field at $x=d_1$ and $x=d_2$ is essentially the same.

So, they might be talking about the symmetry about yz-plane

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