Question

Derive an expression for conductivity of conductor in terms of relaxartion time

Answer 1

conducting solid, an electron will suffer collisions with fixed heavy ions. After collisions,

electron will emerge with same speed, but direction changes randomly. If we consider

N number of electrons in a given volume, since directons are changed randomly due to collisions,

average velocity of N electrons will be zero.

This is expressed as .................(1)

If electrons are accelerated by electric field E, then acceleration is given by, a = -eE/m ..........(2)

 

Let us consider an ith electron in a group of N electrons at a given time t.

Let us assume after a previous collision, speed of this ith electron is vi and there is an elapsed time ti after collision. 

 

Speed Vi of this ith electron at time t is given by,   Vi = vi - (eE/m)ti ...............(3)

 

Average velocity of electrons at time t is average of all Viof each electron in the group we have considered.

In eqn.(3), average of vi appearing on left side is zero as mentioned in eqn.(1).

Collisions of elecrons do not occur at regular intervals but at random time. Let us denote the average

time between successive collisions as τ.

 

Then averaging eqn.(3) over N electrons at any given time t gives us average velocity vd, as given by

vd

............................(4)

vd is called drift velocity. Due to drift, there will be net transfer of charges across any area perpendicular to Electric field E.

 

Consider a planar area A, located inside the conductor such that normal to area is parallel to Electric filed E.

Then because of drift, in an infinitesimal amount of time Δt, all electrons to the left of the area at distances

upto |vd|Δt would have crossed the area. If n is number of free electrons per unit volume in the metal,

then there are nΔt|vd|A such electron. Since each electron carry a charge -e, the total charge transported

across this area A to the right in time Δt is -neA|vd|Δt.

 

Flow of charge per unit time across an area A is the magnitude of current I.


Then we have, I = neA|vd| ..................(5)

by substituting vd from eqn.(4) in eqn.(5),   ..............................(6)

Current density J is defined as, J = I/A, where I is the current flowing in a cross section area A.

 

hence eqn.(6) is written as, J = σE,  where conductivity σ is expressed as