The electric field E, current density J and conductivity $\sigma$ of a conductor are related as

$$\begin{gathered} Resis\tan ce, R =\rho \frac{l}{A} \\ \therefore \rho =\frac{RA}{l} \\ But R=\frac{V}{i} \\ \therefore \rho =\frac{VA}{il}=\frac{V/l}{i/A} \\ But \frac{V}{l}=E and \frac{i}{A}=J \\ \therefore \rho =\frac{E}{J} \\ or E=\rho ·J \\ or E=\frac{1}{\sigma }J \\ where, \sigma =\frac{1}{\rho } \\ Thus, \sigma =J/E \end{gathered}$$