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Tow identical blocks of charge $q$ each are connected by massless spring of force constant $k$ . They are placed over a smooth horizontal surface. They are released when the separation between them is $r$ and spring is unstretched. If maximum extension of the spring is $r,$ the value of $k$ is (neglect gravitational effect)

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$\frac{q}{4 r} \sqrt{\frac{1}{π ε_{0} r}}$

$\frac{q}{2 r} \sqrt{\frac{1}{π ε_{0} r}}$

$\frac{2 q}{r} \sqrt{\frac{1}{π ε_{0} r}}$

None of these

answer is D.

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Detailed Solution

From conservation of mechanical energy

$\frac{1}{2} k r^{2} = \frac{1}{4 π ε_{0}} [\frac{q^{2}}{r} - \frac{q^{2}}{r + r}]$ Or $\frac{1}{2} k r^{2} = \frac{q^{2}}{8 π ε_{0} r}$

$∴ k = \frac{q^{2}}{4 π ε_{0} r^{3}}$

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