A charge +q is distributed over a thin ring of radius r with line charge density \lambda = q\sin^2\theta\/(\pir). Note that the ring is in the x-y plane and θ is the angle made by r^{\to} with the x-axis. The work done by the electric force in displacing a point charge +Q from the center of the ring to infinity is

Correct Answer is : equal to qQ∕4πε0r

equal to zero only if the path is a straight line perpendicular to the plane of the ring.

KVPY 2019 SB SX Exam Previous Papers

Question : 36 of 120

Marks: +1, -0

A charge +q is distributed over a thin ring of radius r with line charge density

λ = q\sin^2θ\/(πr).

Note that the ring is in the x-y plane and θ is the angle made by

r↖{→}

with the x-axis. The work done by the electric force in displacing a point charge +Q from the center of the ring to infinity is

equal to $qQ\/2πε_0r$
equal to $qQ\/4πε_0r$
equal to zero only if the path is a straight line perpendicular to the plane of the ring.
equal to $qQ\/8πε_0r$

Solution:

so small potential at center

dv ={kdq}/ r

dv= k{λ(rdθ)}/r

=k[{q\sin^2θ}/(πr)]dθ

so net potential

v =∫dv={kq}/{πr}∫↖{2π}↙{0}({1-\cos2θ}/2)dθ

v={kq}/{πr}∫↖{2π}↙{0}[1/2θ-{\sin2θ}/4]^{2π}_0

v ={kq}/{ r}=1/{4πε_0}q/r

so work done =

Q [v_i -v_\f] ={qQ}/{4πε_0r}

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