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# 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 qAv2πx3The 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) =ε0KqAdx−2dt =ε0×14πε0×q×A×(−2x−3)dxdt [∵K=14πε0] =−qAv2πx3 [∵v=dxdt] ∴|Id|=qAv2πx3 Hence, option (A) is the correct answer.  Suggest Corrections  0      Similar questions  Explore more