# Why are $\vec{E } \cdot \hat{n}$ and $\vec{J} \cdot \hat{n}$ needed to prove the electric field is uniform in a wire

Here is an excerpt from Griffith's Introduction to Electrodynamics third edition. In the proof, what was the rationale in deciding to do the dot product of $\vec{J}$ with a unit vector and $\vec{E}$ with a unit vector? How does $J\cdot\hat{n}=0$ verify that there is no current density?

• Hi Oscar; please don't post images of text or mathematical formulas. You should copy the relevant part of the text into your post, using MathJax where necessary. (While you're at it, a better title would also be a huge improvement.) – David Z Jan 10 '15 at 1:38

$\hat{n}$ is by convention the unit vector that is normal to a surface, in this case the curved surface of the conductor. So $\hat{n}$ sticks perpendicularly out from every point on the curved surface of the conductor.
Since $\vec{J}$ is the vector field representing the current, if $\vec{J} \cdot \hat{n}$ were nonzero on the surface of the conductor, then the component of $\vec{J}$ in the direction of $\hat{n}$, which is directly out of the surface, would be nonzero, which means that current is moving out of the surface of the conductor.
Griffith points out that by defining one region as the "conductor", we are assuming that the region outside of the conductor is an insulator, which means there should be no movement of charge outside the conductor, so the current flowing out of the curved surface of the conductor should be zero, therefore $\vec{J} \cdot \hat{n}$ must be zero.
All that is just to establish the boundary condition for the current field $\vec{J} \cdot \hat{n} = 0$ on the curved surface, which allows him to find the solution to Laplace's equation for V.
Note that he is not saying that there is no current density, just that the current density is uniform, and in the $\hat{z}$ direction, throughout the conductor.