Question

What is the relationship between electric intensity E and current density J?

Answer 1

The relationship between electric field intensity E and current density J is given by Ohm's law in a more general form for a material, expressed as:

\[
\mathbf{J} = \sigma \mathbf{E}
\]

Where:

1. J is the current density (the amount of electric current flowing per unit area, typically in amperes per square meter, A/m²).
   

1. E is the electric field intensity (the force experienced by a charge in an electric field, measured in volts per meter, V/m).
   

1. σ is the electrical conductivity of the material (a constant that measures how easily a material allows the flow of electric current, measured in siemens per meter, S/m).
   

Explanation:

1. The electric field E causes charges to move, resulting in an electric current. The current density J describes how much current flows through a unit area of a conductor.
   

1. The relationship shows that the current density is directly proportional to the electric field intensity, with the conductivity σ acting as the proportionality constant. This means that in materials with high conductivity (like metals), a smaller electric field is needed to produce a large current density.
   

In simpler terms:

1. Higher electric field (E) means more current (J) for the same material.
   

1. The material's ability to carry current depends on its conductivity. Materials like copper (with high conductivity) will have a larger current density for the same electric field compared to a material like rubber (which has very low conductivity).