    Question

# What is the relation between current density and electric field?

Open in App
Solution

## Given information:Electric field (E) and current density (J).To find:We have to find the relation between current density and electric field.Calculations: We know that,$\mathrm{V}=\mathrm{IR}$and,$\mathrm{R}=\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}$Here,R is the resistance.I is the current.V is the voltage.$\mathrm{\rho }$ is the resistivity of the material.L is the length of the conductor.A is the cross-sectional area.$\therefore \mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{V}}{\mathrm{\rho }\frac{\mathrm{l}}{\mathrm{A}}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{VA}}{\mathrm{\rho l}}$Now,$⇒\frac{\mathrm{I}}{\mathrm{A}}=\frac{1}{\mathrm{\rho }}\frac{\mathrm{V}}{\mathrm{l}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{J}\propto \mathrm{E}$Therefore, the relation between current density and an electric field is $\mathrm{J}\propto \mathrm{E}$.  Suggest Corrections  13      Similar questions  Explore more