# 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 Aev_{d}

I = n Ae(eE/m) ( because v_{d}=eE/m)

I = nAe^{2}E/m

Or I/A=ne^{2} 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/ne^{2}t

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

J=E/ρ

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

I=∫J.ds

= ∫JdS cos 0^{0}

=J∫dS

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

V=JA*l/σA

V=Jl/σ

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.**