Question

A point charge is moving along a straight line with a constant velocity v. Consider a small area A perpendicular to the 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.

Answer 1

From Coulomb's law :
Electric field strength,

`E = (kq)/x^2`

Electric flux,

`phi_E = EA`

`phi_E = (kqA)/x^2`

`"Displacement current" = I_d`

`I_d = |∈_0 (dphi_E)/dt|`

`I_d = |∈_0 d/dt((kqA)/x^2)|`

`I_d = |∈_0 kqA d/dtx^-2|`

`I_d = |∈_0 1/(4pi∈_0) xx q xx A xx (-2)x^-3 xx dx/dt|`

`I_d = |(qAv)/(2pix^3)|`