A slab of insulating material of uniform thickness d, lying between \frac{-d}{2} to \frac{d}{2} along the x axis, extends infinitely in the y and z directions, as shown in the figure.

Given:
- Slab thickness: $d$
- Slab extends infinitely in the $y$ and $z$ directions.
- Slab has a uniform charge density: $\rho$
- Electric field is zero at the middle of the slab, at $x=0$.
To find the behavior of the electric field at the surface of one side of the slab, we can consider a Gaussian surface just outside the slab on one side. The Gaussian surface is a cylindrical surface with radius $r$ and height $h$, centered on the $x$-axis. Let's denote the charge enclosed by this Gaussian surface as $Q_{\text{enc}}$.
By symmetry, the electric field $𝐄$ must be uniform and directed radially outwards from the $x$-axis at every point on the Gaussian surface.
We can apply Gauss's Law to the Gaussian surface to relate the electric field to the enclosed charge:
$\oint 𝐄·d𝐀=\frac{Q_{\text{enc}}}{{\epsilon }_0}$
Since the electric field is constant and parallel to the area vector, the dot product simplifies to $𝐄·𝐀=EA$, where $A$ is the magnitude of the area of the Gaussian surface.
Considering that the only charge enclosed is from the slab, $Q_{\text{enc}}$ is equal to the total charge of the slab, which is given by $Q_{\text{enc}}=\rho ·V$, where $V$ is the volume enclosed by the Gaussian surface.
Now, let's determine the volume $V$ enclosed by the Gaussian surface. The volume can be obtained by multiplying the area of the circular base by the height of the cylinder. Assuming the radius of the circular base is $r$ and the height of the cylinder is $h$, we have $V=\pi r^{2}h$.
Substituting $Q_{\text{enc}}=\rho ·V$ into Gauss's Law, we get:
$EA=\frac{\rho ·V}{{\epsilon }_0}$
Simplifying further:
$E·(\pi r^{2}h)=\frac{\rho ·(\pi r^{2}h)}{{\epsilon }_0}$
Simplifying the equation above, we find:
$E=\frac{\rho }{{\epsilon }_0}$
Therefore, the electric field ${𝐄}_{\overrightarrow{}}$ at the surface of one side of the slab is equal to $\frac{\rho }{{\epsilon }_0}$.
Hence, the correct statement is:
'The electric field ${𝐄}_{\overrightarrow{}}$ at the surface of one side of the slab is $\frac{\rho }{{\epsilon }_0}$.'

The electric field $𝐄$ at the surface of one side of the slab is directed radially outward or inward, perpendicular to the surface.
Explanation:
Since the electric field is zero at the center of the slab ($x=0$), the electric field lines must be symmetric about this point. Due to the infinite extension of the slab in the $y$ and $z$ directions, the electric field lines should also be symmetric in those directions.
Considering a point on the surface of the slab, the electric field lines will be perpendicular to the surface. Since the slab has a uniform charge density, the electric field lines will be radial, either directed outward or inward.
Therefore, the statement that is true is that the electric field $𝐄$ at the surface of one side of the slab is directed radially outward or inward, perpendicular to the surface.

Step 1:
To solve the problem, let's consider a Gaussian surface just outside the surface of the slab on one side. The Gaussian surface is a cylinder of length L and radius R, with its axis perpendicular to the x-axis and centered at x = -d/2, where the slab starts.
The electric field at any point on the Gaussian surface can be written as ${𝐄}_{\text{total}}={𝐄}_{\text{slab}}+{𝐄}_{\text{outside}}$, where ${𝐄}_{\text{slab}}$ is the electric field due to the slab and ${𝐄}_{\text{outside}}$ is the electric field due to the charges outside the slab.
Since the electric field is zero in the middle of the slab, at x = 0, and the slab has a uniform charge density $\rho$, the electric field due to the slab is zero at any point inside the slab. Therefore, ${𝐄}_{\text{slab}}=\mathbf{0}$.
Step 2:
Now, let's calculate ${𝐄}_{\text{outside}}$. The electric field due to the charges outside the slab can be obtained using Gauss's Law. Gauss's Law states that the flux of the electric field through a closed surface is equal to the total charge enclosed divided by the permittivity of free space (${ϵ}_0$).
Since the Gaussian surface is outside the slab, the flux through the surface due to the charges outside the slab is equal to the flux through the surface due to the charges on the slab. Therefore, we can write:
$\Phi =\oint {𝐄}_{\text{total}}·d𝐀=\frac{Q_{\text{enclosed}}}{{ϵ}_0}$
The total charge enclosed within the Gaussian surface is simply the charge density $\rho$ multiplied by the volume of the slab inside the Gaussian surface. The volume of the slab inside the Gaussian surface is the length L of the Gaussian surface multiplied by the thickness d of the slab. Therefore, $Q_{\text{enclosed}}=\rho Ld$.
The flux through the Gaussian surface is given by:
$\Phi =\int {𝐄}_{\text{total}}·d𝐀$
Since the electric field ${𝐄}_{\text{total}}$ is constant and parallel to the surface of the Gaussian cylinder, the dot product ${𝐄}_{\text{total}}·d𝐀$ is simply $E_{\text{total}}dA$. Therefore, we can write:
$\Phi =\int E_{\text{total}}dA=E_{\text{total}}\int dA=E_{\text{total}}A$
where A is the surface area of the Gaussian cylinder, given by $A=2\pi RL$.
Substituting the values into Gauss's Law equation, we have:
$E_{\text{total}}A=\frac{Q_{\text{enclosed}}}{{ϵ}_0}$
$E_{\text{total}}2\pi RL=\frac{\rho Ld}{{ϵ}_0}$
Step 3:
Simplifying the equation, we find:
$E_{\text{total}}=\frac{\rho d}{2{ϵ}_0}$
Therefore, the electric field ${𝐄}_{\text{total}}$ at the surface of one side of the slab is given by:
${𝐄}_{\text{total}}=\frac{\rho <br>d}{2{ϵ}_0}\widehat{𝐧}$
where $\widehat{𝐧}$ is a unit vector normal to the surface of the slab.
In conclusion, the statement that is true of the electric field ${𝐄}_{\text{total}}$ at the surface of one side of the slab is:
${𝐄}_{\text{total}}=\frac{\rho d}{2{ϵ}_0}\widehat{𝐧}$