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Two identical charge particles having charge q and mass m are placed ‘L’ distance apart. If one charge is fixed and other free to move. Find its speed when distance between them becomes 2L.( $K = \frac{1}{4 π ε_{0}}$ )

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$\sqrt{\frac{k q^{2}}{m l}}$

$\sqrt{\frac{2 k q^{2}}{m l}}$

$\sqrt{\frac{k q^{2}}{2 m l}}$

$\sqrt{\frac{k q^{2}}{4 m l}}$

answer is B.

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Detailed Solution

$\begin{array}{l} I n t i a l P E = \frac{q^{2}}{4 π \in_{0} L} \\ F i n a l P E = \frac{q^{2}}{4 π \in_{0} 2 L} \end{array}$

$Δ K E = \frac{1}{2} m v^{2} = \frac{q^{2}}{4 π \in_{0}^{} L} (1 - \frac{1}{2})$

$\begin{array}{l} v = \frac{2 q^{2}}{4 π ε_{0} L m} \frac{1}{2} \\ v = \sqrt{\frac{K q^{2}}{L m}} \end{array}$

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