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A non conducting rod `AB` of length `sqrt(3)R`, uniformly distributed charge of linear charge density `lambda` 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
image
A. `(Qlambda)/(4piepsilon_(0))ln(2+sqrt(3))`
B. `(Qlambda)/(2piepsilon_(0))ln(2+sqrt(3))`
C. `(Qlambda)/(4piepsilon_(0))ln(2-sqrt(3))`
D. `(Qlambda)/(2piepsilon_(0))ln(2-sqrt(3))`

1 Answer

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Best answer
`dU=V_(P("ring"))lambdadx=(kQlambda)/(sqrt(R^(2)+x^(2)))dx`
`U=(Qlambda)/(4piepsilon_(0))int_(0)^(sqrt(3)R)(dx)/(sqrt(R^(2)+x^(2)))`
Let `x=R tan theta`
`rArr dx=R sec^(2)theta d theta`
If `x=0 rArr theta=0`, and `x=sqrt(3)R`
`rArr theta=pi//3`
`U=(Qlambda)/(4piepsilon_(0))int_(0)^(pi//3)sectheta d theta=(Qlambda)/(4piepsilon_(0))[ln(sectheta+tantheta)]_(0)^(pi//3)=(Qlambda)/(4piepsilon_(0))ln(2+sqrt(3))`
image

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