What are the units of conductivity $\mathit{\sigma }$, resistivity $\mathit{\rho }$, resistance $\mathit{R}$, and current density $\mathit{J}$?

The conductivity is $\sigma$.

The resistivity is $\rho$.

The resistance is$R.$

The current density is $J$.

The resistance of the resistor is directly to the length and resistivity and inversely proportional to the area.

The expression to calculate the resistance of the resistor is given as follows

$\mathit{R}\mathbf{=}\mathit{\rho }\frac{\mathbf{L}}{\mathbf{A}}$ …… (i)

Here, is the length and is the area of the resistor.

The expression to calculate the current density is given as follows.

$\mathit{J}\mathbf{=}\frac{\mathbf{I}}{\mathbf{A}}$ …… (ii)

Here, is the current.

The expression to calculate the resistance in terms of voltage and current is given as follows.

$\mathit{R}\mathit{=}\frac{\mathit{V}}{\mathit{I}}$ …… (iii)

Here, $\mathit{V}$is the voltage across the resistor.

The unit of the area is ${\text{m}}^2$, unit of length is $m$, $A$unit of current is , and unit of voltage is $\text{V}$.

Calculate the unit of the resistance.

Substitute$V$ for the voltage unit, $A$for the current unit into equation (iii).

$R=\frac{V}{A}$

R=V/A

The volt per ampere can also be written as ohm $\left(\Omega \right)$.

$R=\Omega$.

Therefore, the unit of resistance is ohm $\left(\Omega \right)$.

Derive the expression for the resistivity from the equation (i).

$\rho =\frac{RA}{L}$

Substitute $\left(\Omega \right)$for resistance unit, ${\text{m}}^2$for area unit and for length unit.

$$\begin{gathered} \rho =\frac{\Omega ·{\text{m}}^2}{\text{m}} \\ \rho =\Omega ·\text{m} \end{gathered}$$

Hence the unit of resistivity is ohm-meter.$\left(\Omega ·\text{m}\right)$

The conductivity is reciprocal of the resistivity.

$\sigma =\frac{1}{\rho }$

Substitute $\Omega ·\text{m}$for $\rho$into above equation.

$$\begin{gathered} \sigma =\frac{1}{\Omega ·\text{m}} \\ \sigma ={\left(\Omega ·\text{m}\right)}^{-1} \end{gathered}$$

Therefore, the unit of the conductivity is ohm-meter inverse

Calculate the unit of the current density.

Substitute for current unit, and for area unit into equation (ii).

$J=\frac{\text{A}}{{\text{m}}^2}$

Therefore, the unit of the current density is ampere per meter square$A/m^2$.

Hence the unit of the conductivity, resistivity, resistance and current density are${\left(\Omega ·\text{m}\right)}^{-1}$ , $\Omega ·\text{m}$, $\Omega$ and $\mathrm{A}/{\mathrm{m}}^2$ respectively.