Question

Consider a cylindrical conductor of length I and area of cross-section A. Current I is maintained in the conductor and electrons drift with velocity v_d `(|vec v_d| = (e|vec E|)/m tau)`,  (where symbols have their usual meanings). Show that the conductivity o of the material of the conductor is given by σ = `(n e^2)/m tau`.

Answer 1

If n = number of free electrons per unit volume, Charge passing per second through area A:

I = nqAv_d

Since electron charge magnitude = e,

I = neAv_d

Current density (J) = `I/A` = neAv_d

Substitute drift velocity:

v_d = `(e E)/m tau`

J = `n e((e E)/m tau)`

= `(n e^2 tau)/m E`

Compare with Ohm’s law (microscopic form):

J = σE

Comparing:

σ = `(n e^2 tau)/m`

Conductivity depends on the number of charge carriers n, relaxation time τ, and electron charge and mass