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Two charges each ‘Q’ are released when the distance between is ‘d’. Then the velocity of each charge of mass ‘m’ each when the distance
between them is ‘2d’ is

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$\frac{Q}{4 \sqrt{{πε}_{0} dm}}$

$\frac{Q}{\sqrt{2 {πε}_{0} dm}}$

$\frac{Q}{\sqrt{4 {πε}_{0} dm}}$

$\frac{Q}{\sqrt{8 {πε}_{0} dm}}$

answer is B.

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Detailed Solution

gain in K.E = loss in P.E

$2 . \frac{1}{2} m v^{2} = \frac{- K Q^{2}}{2 d} - (\frac{- K Q^{2}}{d})$

$v = \frac{Q}{\sqrt{4 π ε_{0} d m}}$

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