# Poynting Vector (Griffiths 8.1)

by Rubiss
Tags: griffiths, poynting, vector
 P: 21 1. The problem statement, all variables and given/known data Problem 8.1 from the third edition of Introduction to Electrodynamics by Griffiths: Calculate the power (energy per unit time) transported down the cables of Ex. 7.13 and Prob. 7.58, assuming the conductors are held at potential difference V, and carry current I (down one and back up the other) 2. Relevant equations $$P=\frac{dW}{dt}=\int_{S}\vec{S} \cdot d\vec{a}=\frac{1}{\mu_{0}}\int_{S}(\vec{E}\times \vec{B}) \cdot d\vec{a}$$ 3. The attempt at a solution I have already solved this problem, but am unsure of some of the technicalities. Here we go: The only region where there is both an electric and magnetic field is in the region ab, there is no magnetic field or electric field. Because the two conductors are held at a potential difference V, the electric field is [tex] \vec{E}=\frac{V}{s \cdot \ln{(\frac{b}{a})}} \hat{s} \mbox{ for a
 P: 2 Isn't the electric field: $\vec{E}$ = $\frac{V}{s\cdot ln(\frac{b}{a})}$$\hat{s}$ ($\ast$) obtained by treating the nested cylinders as a wire and then taking the line integral from a to b of the electric field? $\vec{E}$ = $\frac{\lambda}{2 \pi \epsilon_0 s}$$\hat{s}$ (1) This is the E-Field of a long wire. Calculated by using a Gaussian surface. V = $∫^{b}_{a}$ $\vec{E}$ $\cdot$ d $\vec{l}$ insert equation (1) V = $∫^{b}_{a}$ $\frac{\lambda}{2 \pi \epsilon_0 s}$ds V = $\frac{\lambda ln(\frac{b}{a})}{2 \pi \epsilon_0}$ (2) Now rewrite equation (1) with equation (2): $\vec{E}$ = $\frac{V}{s\cdot ln(\frac{b}{a})}$$\hat{s}$ This is the only way I know how to get ($\ast$). Please let me know how you calculated it. (Sorry for the lack of LaTeX skills!)