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 . 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.
(##HCV_VOL2_C40_E01_023_Q01##)

Explanation

Electric field lines of a point charge are radially outwards or inwards depending on whether charge is positive or negative .SO direction of electric field at a point on line of motion of point charge will be along the line of motion and hence perpendicular to the area selected . hence magnitude of the electric flux associated with this area when charge is at a distance x from area can be written as follows :
` phi - ( qA)/( 4 pi epsi_0 x^2)`
Note that above electric flux is written for every small area A. Displacement current through any surface can be written as :
`i_d = epsi_0 (d phi )/( dt) `
`=I_d = (qA d ( x^(-2) ) )/( 4 pi d t)`
` = I_d = (qA )/( 4 pi ) (d (x^(-2) ) /( dx )) ((dx )/( dt ))`
Hence , magnitude of displacement current can be written as :
` implies i_d = (qA )/( 4 pi ) (|-2| )/(x^3)_ (v) `
` implies I_d = (qA v )/( 2 pi x ^3 )`
We can see that displacement current is inversely proportional to the cube of the distance of point charge from that point .