A dipole of dipole moment p is kept at origin, along x direction and fixed. A bead of mass m, charge q can slide on fixed smooth rod kept along line x = c. Bead is given a gentle push so that it starts moving parallel to +y direction from point P as shown. Whole arrangement is kept in gravity free space. [c is very large in comparision to length of dipole]. Then

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speed of bead where normal force between rod and bead is zero, is $\sqrt{\frac{p q}{2 π \in_{0} m c^{2}} [1 - \frac{1}{3 \sqrt{3}}]}$

co-ordinates of point where normal force between bead and rod is zero are $(c, \sqrt{2} c)$

acceleration of bead where normal force between rod and bead is zero is $\frac{p q}{12 \sqrt{3} π \in_{0} m c^{3}}$

normal force between bead and rod at point P is $\frac{p q}{2 π \in_{0} c^{3}}$

answer is A, B, D.

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Detailed Solution

Normal force is zero where E due to dipole is paralle to y-direction.
Question Image
For A : y-direction
$\tan α = \frac{1}{2} \tan (θ) a n d α + θ = \frac{π}{2}$
$\Rightarrow \tan θ = \sqrt{2}$
So, co-ordinates are $(c, \sqrt{2} c)$
For B : $\frac{1}{2} m v^{2} = q [\frac{k p}{c^{2}} - \frac{k p \cos θ}{[c^{2} + 2 c^{2}]}] = \frac{k p q}{c^{2}} [1 - \frac{1}{3 \sqrt{3}}]$
$v = \sqrt{\frac{p q}{2 π \in_{0} m c^{2}} [1 - \frac{1}{3 \sqrt{3}}]}$
For C : $E = \frac{k p}{3 \sqrt{3} c^{3}} \sqrt{1 + \frac{3}{3}} = \frac{\sqrt{2}}{3 \sqrt{3}} \frac{k p}{c^{3}}$
$a c c . = \frac{E q}{m} = \frac{\sqrt{2} p q}{12 π \in_{0} \sqrt{3} c^{3} m}$

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