InterviewSolution
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InterviewSolution
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A non-polar molecule is located at the axis of a thin uniformly charged ring of rasius R. At what distancex fromthe ring's centre is the magnitude of the force F acting on the given molecule (a) equal tozero, (b) maximum ? Draw the appoximate plot F_(x) (x). |
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Answer» Solution :The electric field`E`at distance `x` from the centre of the ring is, `E(x)= (qx)/(4pi epsilon_(0) (R^(2) + x^(2))^(3//2))` The induced DIPOLE moment is `p = beta epsilon_(0) E = (q beta x)/(4pi (R^(2) + x^(2))^(3//2))` The force on this molecule is The force on this molecule is `F = p (del)/(del x) E = (q beta x)/(4pi (R^(2) + x^(2))^(3//2))(q)/(4pi epsilon_(0)) (del)/(dx) (x)/((R^(2) + x^(2))^(3//2)) = (q^(2) beta)/(16 pi^(2) epsilon_(0)) (x (R^(2) - 2x^(2)))/((R^(2) + x^(2))^(4))` Thisvanishes for `x = (+-R)/(sqrt(2))` (apart from `x = 0, x = oo`) It is maximum when `(del)/(del x) (x (R^(2) - x^(2) xx 2))/((R^(2) + x^(2))^(4)) = 0` or, `(R^(2) - 2x^(2)) (R^(2) + x^(2)) - 4x^(2) (R^(2) + x^(2)) - 8X^(2) (R^(2) - 2x^(2)) = 0` or, `R^(4) - 13 x^(2) R^(2) + 10 x^(4) = 0`, or `x^(2) = (R^(2))/(20) (13 +- sqrt(129))` or, `x = (R )/(sqrt(20)) sqrt(13 +- sqrt(129))` (on either side), Plot of `F_(n) (x)` is as shown in the ansersheet. |
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