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?
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.
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?
No comments:
Post a Comment