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Four identical point charges are fixed at the four corners of a square of a side length l. Another charged particle of mass m and charge +q is projected towards centre of square from a large distance along the line perpendicular to plane of square. The minimum value of initial velocity V₀ in m/s required to cross the square is? $(m = 1 gm, l = 4 \sqrt{2} m, Q = 1 μc, q = 0 .5 μc)$

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answer is 3.

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Detailed Solution

As the charge approaches the potential energy of the system of particles will increases.

Initially : Kinetic energy = $\frac{1}{2} {mv}_{o}^{2}$ , Potential energy = 0

At the moment when the particle just crosses the centre of square.

Kinetic energy = 0 , Potential energy = $4 . \frac{KQq}{\frac{l}{\sqrt{2}}}$

$\Rightarrow \frac{1}{2} {mv}_{o}^{2} = \frac{4 KQq}{\frac{l}{\sqrt{2}}}$

$\Rightarrow \frac{1}{2} \times 10^{- 3} \times {v_{o}}^{2} = \frac{4 \times 9 \times 10^{9} \times 10^{- 6} \times 0.5 \times 10^{- 6}}{4}$

$\Rightarrow v_{o} = 3 m / s$

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