- #1
stephen8686
- 42
- 5
Homework Statement
If the magnitude of the drift velocity of free electrons in a copper wire is 7.84×10-4 m/s, what is the electric field in the conductor? (Also gives chart that states that the resistivity of Cu is 1.7×10-8 Ωm)
Homework Equations
[/B]
vd=(qEτ)/me (where τ is avg. time between collisions)
ρ=1/σ=me/(nq2τ)
The Attempt at a Solution
first I solved for τ in the second equation:
τ=me/(nq2ρ)
Then I put this value in for τ in the first equation:
vd=(qE(me/(nq2ρ))/me
Simplify:
vd=E/(qnp)
When I put in values:
7.84×10-4=E/(1.6×10-19)(1.7×10-8)(n)
I didn't know how to get n (number of e- per m3), so I googled the density of copper (8.96 g/ml) and did some conversions to get n=8.475×1028, which gave me a correct answer of E=.18V/m, but I don't think I was supposed to use google, and no density chart is given in the book. Is there another way to do this that I overlooked?