Question

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 its natural length. If maximum extension of the spring is r, the value of $\sqrt{K}$ is (Neglect gravitational effect)

• A. $\frac{q}{4r}\sqrt{\frac{1}{\pi {\epsilon }_{0}r}}$
• B. $\frac{q}{2r}\sqrt{\frac{1}{\pi {\epsilon }_{0}r}}$
• C. $\frac{2q}{r}\sqrt{\frac{1}{\pi {\epsilon }_{0}r}}$
• D. $\frac{q}{r}\sqrt{\frac{1}{\pi {\epsilon }_{0}r}}$

Answer 1

The correct option is B $\frac{q}{2r}\sqrt{\frac{1}{\pi {\epsilon }_{0}r}}$

From conservation of mechanical energy $\frac{1}{2}K{r}^2=\frac{1}{4\pi {\epsilon }_0}[\frac{q^2}{r}-\frac{q^2}{r+r}]or\frac{1}{2}K{r}^2=\frac{q^2}{8\pi {\epsilon }_{0}r}\therefore \sqrt{K}=\frac{q}{2r}\sqrt{\frac{1}{\pi {\epsilon }_{0}r}}$