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Two identical particles of charge q each are connected by a mass less 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 in it’s natural length. If maximum extension of the spring is r, the value of K is (neglect gravitational effect)

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$\frac{q}{2 r} \sqrt{\frac{1}{{πε}_{0} r}}$

$\frac{q}{4 r} \sqrt{\frac{1}{{πε}_{0} r}}$

$\frac{2 q}{r} \sqrt{\frac{1}{{πε}_{0} r}}$

$\frac{q}{r} \sqrt{\frac{1}{{πε}_{0} r}}$

answer is B.

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Detailed Solution

After an elongation of r, the separation between the charges will be '2r'

$\begin{array}{l} \frac{1}{4 {πϵ}_{0}} \times \frac{q^{2}}{2 r} = \frac{1}{2} {kr}^{2} \\ k = \sqrt{\frac{q^{2}}{4 {πϵ}_{0} r^{3}}} = \frac{q}{2 r} \sqrt{\frac{1}{{πϵ}_{0} r}} \end{array}$

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