Alternate Answer
Conductivity is defined as the reciprocal of resistvity of a metallic wire.
S.I. unit
$\Omega^{-1}\text{m}^{-1}\ \text{or mho (m}^{-1}).$$\rho=\frac{\text{RA}}{\text{l}}$
but $\text{R}=\frac{\text{ml}}{\text{ne}^2\text{A}\tau}$
So, $\rho=\Big(\frac{\text{ml}}{\text{ne}^2\text{A}\tau}\Big)\frac{\text{A}}{\text{l}}$
$\rho=\frac{\text{m}}{\text{ne}^2\tau}\ \text{or}\ \sigma=\frac{\text{ne}^2\tau}{\text{m}}\ ....(1)$
where $\sigma=\frac{1}{\rho}=$ conductivity
$\tau=$ relaxation time.
n = number density of electron.
e = charge on one electron.
m = mass of one elecrtron.
Now, current density is given by:
$\text{J}=\text{neV}_\text{d}$
$\text{J}=\text{ne}\Big(\frac{\text{eE}\tau}{\text{m}}\Big)$ $\Big[\because\ \text{V}_{\text{d}}=\frac{\text{eE}\tau}{\text{m}}\Big]$
$\text{J}=\frac{\text{ne}^2\tau}{\text{m}}\text{E}$
From equation (1)
$\text{J}=\sigma\text{E}\ \text{or }\ \vec{\text{J}}=\sigma\vec{\text{E}}$
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