conductorselectric-fieldselectrostaticsgauss-lawhomework-and-exercises
The field near the surface of a conductor is given by
$$\bf{E}=\frac{\sigma}{\epsilon_0} \hat{\bf{n}}$$
Where $\sigma$ is the local surface charge density.
My question is: is this field $\bf{E}$ always outgoing from the conductor surface (as in picture)? What is the reason why it must be outgoing and it cannot be ingoing in the conductor?
When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes into account all the charge on both these surfaces, so it is the sum of the two charge densities.
$$\sigma = \frac{Q}{A} = \sigma_\text{inside} + \sigma_\text{outside}$$
With this definition, the equation we get from Gauss's law is
$$E_\text{inside} + E_\text{outside} = \frac{\sigma}{\epsilon_0}$$
where "inside" and "outside" designate the regions on opposite sides of the plate. For an isolated plate, $E_\text{inside} = E_\text{outside}$ and thus the electric field is everywhere $\frac{\sigma}{2\epsilon_0}$.
Now, if another, oppositely charge plate is brought nearby to form a parallel plate capacitor, the electric field in the outside region (A in the images below) will fall to essentially zero, and that means
$$E_\text{inside} = \frac{\sigma}{\epsilon_0}$$
There are two ways to explain this:
The simple explanation is that in the outside region, the electric fields from the two plates cancel out. This explanation, which is often presented in introductory textbooks, assumes that the internal structure of the plates can be ignored (i.e. infinitely thin plates) and exploits the principle of superposition.
The more realistic explanation is that essentially all of the charge on each plate migrates to the inside surface. This charge, of area density $\sigma$, is producing an electric field in only one direction, which will accordingly have strength $\frac{\sigma}{\epsilon_0}$. But when using this explanation, you do not also superpose the electric field produced by charge on the inside surface of the other plate. Those other charges are the terminators for the same electric field lines produced by the charges on this plate; they're not producing a separate contribution to the electric field of their own.
Either way, it's not true that $\lim_{d\to 0} E = \frac{2\sigma}{\epsilon_0}$.
The E field exactly on the surface in fact should be undefined, because there are surface charges.
But the E field is well-defined if you remove a small disk from the surface.
Let's call the E field due to the disk be $E_\text{disk}$ and the E field due to the other surface charges be $E_\text{other}$.
Then just above the surface
$$E_\text{disk}=\sigma/2\epsilon_0$$
Just below the surface
$$E_\text{disk}=-\sigma/2\epsilon_0$$
And on the surface, $E_\text{disk}$ is undefined.
Now it is clear that $E_\text{other}$ is smooth across the surface and well-defined on the surface.
And because just above the surface
$$E_\text{other}+E_\text{disk}=\sigma/\epsilon_0$$
and just below the surface
$$E_\text{other}+E_\text{disk}=0$$
it can deduced that
$$E_\text{other}=\sigma/2\epsilon_0$$
$E_\text{other}$ on the surface is hence $\sigma/2\epsilon_0$.
So the "E field on the surface" is in fact $E_\text{other}$, viz., the E field at the surface if you remove a small disk of surface charges from the surface, and is well-defined. It is also the E field experienced by that small disk of surface charges.
Best Answer
As @AccidentalFourierTransform pointed out in the comments, in general the direction of the electric field is related to the sign of the charge (density) generating it:
So the answer to your question is: "depends on the sign of the surface charge on the conductor". In the image you posted, as you can see, the surface charge density is positive and this leads to an electric field pointing away from the surface.
Just to be clear, as you ask whether E can be ingoing in the conductor: if you mean, as I suppose, towards the conductor, then the answer is above; if you on the other hand mean inside the conductor, then the answer is no: there cannot be an electrif field inside a conductor (as pointed out in the image in your question).