E-Field Through Copper Wire given drift velocity

[stephen8686]

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

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v_d=(qEτ)/m_e (where τ is avg. time between collisions)

ρ=1/σ=m_e/(nq²τ)

The Attempt at a Solution

first I solved for τ in the second equation:
τ=m_e/(nq²ρ)

Then I put this value in for τ in the first equation:
v_d=(qE(m_e/(nq²ρ))/m_e
Simplify:
v_d=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?

 

[kuruman]

Is there another way to do this that I overlooked?

There is a more direct way that does not involve τ. The current density is related to the electric field by J = (1/ρ) E and to the drift velocity by J = n e v_d. You still need n which you cannot get from what is given unless you look things up.

 

[stephen8686]

There is a more direct way that does not involve τ. The current density is related to the electric field by J = (1/ρ) E and to the drift velocity by J = n e v_d. You still need n which you cannot get from what is given unless you look things up.

Thanks kuruman. Just wanted to make sure I wasn't missing anything important.