Question

What is the relationship between current density and electric field Class 12?

Answer 1

The relationship between current density and electric field is described by Ohm's Law in its microscopic form. This relationship is given by:

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

Where:

1. \(\vec{J}\) is the current density (measured in amperes per square meter, A/m²),
   

1. \(\vec{E}\) is the electric field (measured in volts per meter, V/m),
   

1. \(\sigma\) is the conductivity of the material (measured in siemens per meter, S/m).
   

Explanation:

1. Current density (\(\vec{J}\)) represents the amount of electric current flowing through a unit area of a material, and it is a vector quantity, meaning it has both magnitude and direction.
   

1. Electric field (\(\vec{E}\)) is the force per unit charge that pushes electrons to move through the material.
   

1. Conductivity (\(\sigma\)) is a property of the material that tells us how easily it allows the flow of electric current. Materials with high conductivity (like metals) allow more current to flow for the same electric field, while materials with low conductivity (like insulators) resist current flow.
   

What this relationship means:

1. The current density \(\vec{J}\) is directly proportional to the electric field \(\vec{E}\). This means that the stronger the electric field applied to the material, the greater the current density.
   

1. The proportionality constant \(\sigma\) depends on the material. For conductors like copper, the conductivity is high, so the current density is large for a given electric field.
   

In simpler terms:

When an electric field is applied across a material, it causes the charges (typically electrons) to move. The faster they move (which leads to more current), the higher the current density. The relationship is linear: doubling the electric field will double the current density, provided the material’s conductivity remains constant.

Does that make sense?