Proof that $\vec{E}$-field is constant inside cylindrical resistor

Question

I am reading a proof that the $\vec{E}$-field is constant inside a cylindrical resistor, and I don't understand one of the steps. It is stated that since the surrounding medium is non-conductive the flow of charge at the surface has no component along the normal of the surface. From this the conclusion is drawn that the $\vec{E}$-field along the normal must be zero too.

This I don't understand. Since the conductivity of the surrounding medium is assumed to approach zero couldn't the $\vec{E}$-field be nonzero without causing charge to flow?

Answer 1

I think it would be better to include the actual figure of the resistor in question. I will do that below (I have also added the normal vector the example is referring to):

Since there is no current along the $\hat n$ direction, it must be that $\mathbf J\cdot\hat {\mathbf n}=0$, and since $\mathbf E$ is proportional to $\mathbf J$, it must be that $\mathbf E\cdot\hat {\mathbf n}=0$ as well.

One issue you are having seems to be with the specification that the surrounding medium is non-conducting. This is specified as an argument for why $\mathbf J\cdot\hat {\mathbf n}=0$ is true, not for why $\mathbf E\cdot\hat {\mathbf n}=0$ is true.

Since the conductivity of the surrounding medium is assumed to approach zero couldn't the $\vec E$-field be nonzero without causing charge to flow?

I suppose you are right here, but, the example is concerned with the field just inside the resistor, since we are wanting to solve Laplace's equation inside that region of space. What the field is doing outside is of no concern to us here (the footnote somewhat discusses the field outside).