Question

derive an expression for electrical conductivity​

Answer 1

metals, the electrical conductivity depends on the number of charge carriers (free electrons) present in the material.

Let us consider a solid metal (S) of length 'l' and area of cross-section 'A'. Let 'n' be the number of electrons per unit volume. Then,

Total number of electrons in the metal N = nAl ----(1)

We know,

Total charge Q = Total number of electrons in the metal x charge of one electron

∴Total charge Q = N(-e) ----(2)

The negative sign indicates that the charge of the electron is negative.

Substituting equation (1) in (2) we get,

Total charge present in the solid Q = nAl(-e)

Now, when we apply a voltage 'V' to the metal, then the electrons in it starts to move with an average velocity called drift velocity (Vd) from one end to another end (through a distance l), giving rise to current conduction in the metal.

∴The current through the metal

Substituting equation (3) in (4) we get,

I =

nAl(-e)

τ

........ (5)

d The current density i.e., the current flowing through the metal per unit area is given by

J =

I

A

........ (6)

Substituting equation (5) in equation (6), we get

J =

nAl(-e)

τ A

J =

nl(-e)

τ

........ (7)

We know, drift velocity

Vd =

Average distance travelled by the electron

Time taken

Vd =

l

τ

........ (8)

Substituting equation (8) in equation (7), we get

Current density J = nVd(-e) .....(9)

Due to the applied electric field, the electrons gain the acceleration 'a'

Acceleration (a) =

Drift velocity (Vd)

Relaxation time (τ)

(or)

Vd = aτ .....(10)

If E is the field intensity and m is the mass of the electron, then the force experienced by the electron is,

F = -eE .......(11)

From Newton's second law,

The force experienced by the electron is, F = ma ..........(12)

Comparing equation (10) and (11) we have

-eE = ma

(or)

a =

-eE

m

........ (13)

Substituting equation (13) in (10) we have

Vd =

eE

m

τ ........ (14)

Substituting equation (14) in (9) we have

J = n(-e)

eE

m

τ

J =

ne2Eτ

m

........ (15)

Here the number of electrons flowing per second through unit area (the current density), depends on the applied field. Thus if the electric field (E) applied is more, current density (J) will also be more.

So, we can write J ᾳ E

or J = σE .....(16)

Comparing equation (15) and (16) we can write

σE =

ne2Eτ

m

σ =

ne2τ

m

Definition for C-efficient of Electrical Conductivity (σ)

It is defined as the quantity of electricity flowing per unit area per unit time maintained at unit potential gradient.

Unit : Ω-1m-1

Answer 2

Answer:

Expression for electrical conductivity​ is given by:

$\sigma= \frac{\big n\big e^2\big \tau}{\big m}$

Explanation:

• Conductivity is the reciprocal of resistivity of the material. It is denoted by σ.

Consider that a conductor of length 'l' and area of cross section A. Let 'n' is the number of electrons in per unit area of the conductor. If electric field E is applied cross the two ends of the conductor the drift velocity is given by:

$V_{d}=\frac{\big e\big E}{\big m}\tau$

The current through the conductor due to the drift velocity is given by:

$I=nAV_{d}.{ \big e}$

Substitute the value of drift velocity in above equation,

$I=nA\big (\frac{\big e\big E}{\big m} \tau \big )e$                                                                .................(1)

If V is the potential difference between two ends of the conductor:

$E=\frac{\big V}{\big l}$

Substitute the value of electric field 'E' in equation (1);

$I=\frac{nAe^2V\tau}{ml}$

$\frac{\big V}{\big I} =\frac{\big m\big l}{\big n\big A\big e^2\tau}$

According to Ohm's law  $\frac{V}{I} =R$, the resistance of the conductor,

$R =\frac{\big m}{\big n\big e^2\tau}.\frac{\big l}{\big A}$                                                                 .............(2)            

Resistance,  $R=\rho \frac{\big l}{\big A}$                                                    ..............(3)

The resistivity of the material of the conductor, by comparing equation (2) and (3):

$\rho =\frac{\big m}{\big n\big e^2\tau}$

Now the conductivity will be: $\sigma=\frac{1}{\rho}$

$\sigma= \frac{\big n\big e^2\big \tau}{\big m}$

Hence, the above equation is an expression for electrical conductivity.