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Two rectangular plates A and B placed at a distance 2a apart, are connected to a battery to produce an electric field. There are insulators between plates C and other two plates. A magnetic field exists along z-axis. A charged particle of mass m and charge q passes through a hole at the middle of the plate A with velocity v and strikes at Q which is the middle of the bottom edge of plate B after passing through a hole in plate C. If E=m v^{2} / q a what will be the speed of the particle at Q?

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$v \sqrt{5}$

$v \sqrt{2}$

$v \sqrt{3}$

2 v

answer is C.

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Detailed Solution

$Work done = q E . 2 a = change in KE q \cdot \frac{m v^{2}}{q a} \cdot 2 a = \frac{1}{2} m v_{Q}^{2} - \frac{1}{2} m v^{2} 4 v^{2} = v_{Q}^{2} - v^{2} \Rightarrow v_{Q} = v \sqrt{5}$

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