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A particle of mass m carrying charge q is projected with velocity v from point P toward an infinite line of charge from a distance a.Its speed reduces to zero momentarily at point Q, which is at a distance a/2 from the line of charge. If another particle with mass m and charge -q is projected with the same velocity v from P toward the line of charge, what will be its speed at Q?

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$2 v$

$\sqrt{2} v$

$\sqrt{v}$

$\sqrt{2}$

answer is A.

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Detailed Solution

$(0 - \frac{1}{2} m v^{2}) = q (V_{P} - V_{Q}) ............... (i)$

$\frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v^{2} = - q (V_{P} - V_{Q}) .............. (i i)$

From Eqs. (i) and (ii),

$v_{1} = \sqrt{2} v$

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