A point charge is moving along a straight line with a constant

velocity v. Consider a small area A perpendicular to the direction of motion of the charge figure. Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large so that the electric field at any instant is essentially given by Coulomb’s law.

Given: Velocity of charge = v

Area of patch = A

Distance of charge from the patch=x

We have to find the displacement current through the area when it is at a distance x from the charge.

The displacement current arises due to changing electric field which in turn produces a varying electric flux through an area. The displacement current depends on the electric flux as

where I_d is the displacement current,ϕ_E is the varying electric flux

through the area and ϵ₀ is the electric permittivity of free

space(vacuum) and is equal to 8.85 × 10^-12 C² N^-1 m^-2.

The electric field produced by the charge when it is at a distance x

from the surface is given by Coulumb’s law and is equal to

This electric field produces an electric flux through the area whose

magnitude is given by Gauss’s law

. This is because the electric field lines are directed along the normal to the area vector of the surface.

The angle θ between them is 0^° so cos 0^° =1.

The displacement current will be

As x is the distance of the charge from the area at different intervals of time so the rate at which the particle is changing its position is its velocity v

Thus the displacement current through the area is