Derive an expression for electrical conductivity.

Explanation

Let .n. be the number density of electrons. Let .L. be the length of the conductor, .A. be the area of cross-section of the conductor.
Let `v_(d)` be the drift velocity of the electrons. Let `Deltax` be a small length.
Electrons drift in a direction opposite to the electric field. Number of electrons in`. Deltax.` and area of cross-section .A. is equal to `(ADeltax)n`.
Charge on these electrons = `(nADeltax)e`

If `Deltat` is the time taken for effective displacement of electrons then rate of flow charge = `nAe((Deltax)/(Deltat))`. By definition, electric current I = rate of flow of charge.
i.e., `I=nAev_(d)` ......(1)
where, average velocity of electrons with which it drifts against the direction of electric field is known as drift velocity `(v_(d))`.
Let .a. be the acceleration of electrons. Let .E. be the electric field intensity. Force on electrons,
F=ma
i.e., eE=ma
but= a= `vtaut` where `.tau.` is the relaxation time. Relaxation time represents the average time taken for two successive collisions of electrons and ions in the lattice.
Hence, `eE=mv_(d)tau` or `v_(d)=(eE)/(mtau)` ...(2)
We also know that electric potential difference between the ends of the conductor V = EL
`v_(d)=(eV)/(mtauL)` substituting this in the expression (1)
We write,
`I=nAe((eV)/(mtauL))`
i.e., `I=((nAe^(2))/(mtauL))V`
`R=(mtauL)/(mAe^(2))` is called the electric resistance of a conductor.
and `K=((nAe^(2))/(mtauL))` is called the electric conductance of a conductor.
the expression `sigma=(nc^(2))/(mtau)` is called the electrical conductivity and `1/(sigma)=(mtau)/(ne^(2))` is called electrical resistivity.