# 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?

## 1 Answer

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).

• So if I understand it correctly he technically isn't making any statement about the E field at the actual surface. However for any point inside the cylinder we can look at a cylinder which contains that point and reaches very close to the boundary. For this cylinder we do have the boundary conditions so we can solve for the internal field. Is that correct? – fibo11235 Nov 6 '18 at 14:37
• @fibo11235 Right! We are concerned with what is happening inside the cylinder, not outside of it – Aaron Stevens Nov 6 '18 at 14:45