Two non conducting circular rings A and B, each of radius a = 30 cm are placed coaxially with their axes horizontal in a uniform electric field $E = 10^{5} N C^{- 1}$ directed vertically upwards as shown in figure. Distance centres of these rings A and B is h = 40 cm. Ring A has a positive charge $q_{1} = 10 μ C$ while ring B has a negative charge of magnitude $q_{2} = 20 μ C .$ A particle of mass m = 100 gm and carrying a positive charge $q = 10 μ C$ is released from rest at the centre of the ring A. The velocity when it has moved a distance of 40 cm is $(g = 10 m / s^{2})$

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$6 \sqrt{7} m / s$

6 m/s

$6 \sqrt{2} m / s$

$6 \sqrt{5} m / s$

answer is B.

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Detailed Solution

The weight of particle is balanced by external electric field force.
$U_{1} = U_{0} + \frac{K q q_{1}}{a} \frac{K q (- q_{2})}{\sqrt{a^{2} + h^{2}}}$
$U_{2} = U_{0} + \frac{K q q_{1}}{\sqrt{a^{2} + h^{2}}} + \frac{K q (- q_{2})}{a}$
$U_{1} = U_{2} + \frac{1}{2} m v^{2}$
$v = 6 \sqrt{2} m / s$