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Question

A non conducting rod AB of length 3R, uniformly distributed charge of linear charge density λ and.a non-conducting ring of uniformly distributed charge Q, are placed as shown in the figure. Point A is the centre of ring and line AB is the axis of the ring, perpendicular to plane of ring. The electrostatic interaction energy between ring and rod is
329404_b8e2b6825d134f0fb48387ac7b5f3183.png

A
Qλ4πϵoln(2+3)
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B
Qλ2πϵoln(2+3)
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C
Qλ4πϵoln(23)
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D
Qλ2πϵoln(23)
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Solution

The correct option is A Qλ4πϵoln(2+3)
Given a non-conducting rod of length 3R; λ= linear charge density;
non-conducting ring having charge =Q
To find electrostatic interation energy between ring and rod.



Small change in energy, dU= potential × charge
=Vp (ring) λdx ...(i)
potential due to ring, Vp (ring) =kQR2+x2 ...(ii)
where, k=14πε0
x= distance of point P from center of ring.
So, dU=kQR2+x2λdx
[Here we have substituted (ii) in (i)]

So, total electrostatic energy can be obtained by integrating 3R
U=Qλ4πε03R0dxR2+x2
To solve this integral, x=Rtanθ
On differentiating both sides, we get
dx=Rsec2θdθ
For limits of integration: if x=0θ=0
and for, x=3Rθ=π3

So, U becomes, U=Qλ4πε0π/30sec2θdθ
=Qλ4πε0[ln(secθ+tanθ)]π/30
=Qλ4πε0ln(2+3)

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