Relation between Current density, conductivity and electric field: Point Form of Ohm’s law

As we have already derived and discussed the Ohm’s law. Let us derive and discuss the point form of Ohm’s law which is basically the relation between current density (J), conductivity (σ) and electric field (E).

As current  is related with drift velocity by relation

I=n Aevd

I = n Ae(eE/m)              ( because vd=eE/m)

I   = nAe2E/m

Or                  I/A=ne2 E/m                            (1)

As current density is given by J=I/A (Refer article of Current Density, Conductance and Electrical Conductivity)

and  resistivity is given by  ρ=m/ne2t

By putting above values of J and ρ in equation (1), we get


Or                                               J=σ E                                      (where conductivity σ =1/p)

Alternate Method:

According to Ohm’s law, the potential difference between ends of a conductor is directly proportional to the current, that is

V α I

Or                                      V=IR                                             (2)

Where  R is resistance (that is obstruction offered to flow of changes)

Also, when a current I flows in a uniform cross-section area normally to it, then


= ∫JdS cos 00


I=JA          (3)     (where ∫dS = A)

Also, resistance                                            R= ρl/A                                                 (4)

Equation (3) can also be written as

R=l/σA                                             (5)

Where σ=1/ ρ is conductivity.

Substituting equations (3) and (5) in equation (2) we get



Or                                J=σV/l                                       (6)

But electric field intensity E=V/l

By putting above value of E in equation (6),we get

J= σ E

This is also known as point form of Ohm’s Law.

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