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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.
Solution
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)|`
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