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

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? 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?