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A positively charged particle of charge q and mass m is suspended from a point by a string of length $l .$ In the space a uniform horizontal electric field E exists. The particle is drawn aside so that the string becomes vertical and then it is projected horizontally with velocity v such that the particle starts to move along a circle with same constant speed v. Find the speed v.

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$\frac{2 q E}{m} \sqrt{\frac{l}{g}}$

$\frac{q E}{2 m} \sqrt{\frac{l}{g}}$

$\frac{q E}{m} \sqrt{\frac{2 l}{g}}$

$\frac{q E}{m} \sqrt{\frac{l}{g}}$

answer is B.

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Detailed Solution

$θ$ is equilibrium position. Net force in equilibrium position

$F = \sqrt{{(m g)}^{2} + {(q E)}^{2}}$

$\tan θ = \frac{q E}{m g}$

$T \cos θ = F$

$T \sin θ = \frac{m v^{2}}{r}$

where $r = l \sin θ$

Solve to get $v = \frac{q E}{m} \sqrt{\frac{l}{g}}$

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