homework and exercises - Electric field outside wire with stationary current

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?