Consider a conductor of arbitrary structure where a **stationary** current flow, that is 

$$\nabla \cdot \vec{j}=0$$

I did not find on textbook clear explanations on two facts:

* How is the **electric field** outside the conductor made exactly? (in particular the normal component).
* Are there charge densities on the surface of the conductor?

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I know that, inside the conductor $$\vec{j} = \sigma \vec{E}$$

($\sigma$ is conductivity). And that the tangential component of $E$ is always conserved.

Moreover, since the normal component of $\vec{j}$ is conserved (this follows from $\nabla \cdot \vec{j}=0$), and $\sigma\_{ext}=0$ we have that
$$j\_{n,int}=j\_{n,ext}=\sigma\_{int}E\_{n,int}=\sigma\_{ext}E\_{n,ext}=0$$

That is, there is no normal current on the surface.

But what is $E\_{n,ext}$ ? I cannot conclude nothing from here.

If there were surface density of charges (named $\varsigma$), then is would be
$$E\_{n,ext}=\frac{\varsigma}{\epsilon\_0}$$
But I'm not sure about the presence of charges. 

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To sum up in my view it should be
$$\begin{cases} E\_{t,ext}=E\_{t,int}=E\_{int}=\frac{j}{\sigma} \\ E\_{n,ext}=\frac{\varsigma}{\epsilon\_0} \\ E\_{n,int}=0 \end{cases}$$

Is this correct or are there any mistakes?