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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 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 too large so that the electric field at any instant is essentially given by Coulomb's law.


A
qAv2πx3
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B
qAvπx3
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C
2qAvπx3
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D
qAvx3
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Solution

The correct option is A qAv2πx3
The electric field due to the point charge q at a distance x is given by,

E=Kqx2

Since, the area A is small we can assume that the electric field is constant and perpendicular to the plane of area A at every point.

Therefore, the electric flux due to this field is given by,

ϕE=EA=KqAx2

Hence, the displacement current through area A is given by

Id=ε0dϕEdt=ε0ddt(KqAx2)

=ε0KqAdx2dt

=ε0×14πε0×q×A×(2x3)dxdt [K=14πε0]

=qAv2πx3 [v=dxdt]

|Id|=qAv2πx3

Hence, option (A) is the correct answer.

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