Symmetry operations on an infinite uniform sheet of charge 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?
 A: Scale symmetry. An infinite plane looks the same no matter how far away from it you are.
A: 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 $E_x$
A: 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
