# Two identical particles of charge q each are connected by a massless spring of force constant K…

Q: Two identical particles of charge q each are connected by a massless spring of force constant K. They are placed over a smooth horizontal surface, Fig. They are released when separation between them is r and the spring is unstretched. If maximum extension of the spring is r , then value of K ( neglecting gravitational effects) is

(a) $\displaystyle \frac{q}{4 r}\sqrt{\frac{1}{\pi \epsilon_0 r}}$

(b) $\displaystyle \frac{q}{2 r}\sqrt{\frac{1}{\pi \epsilon_0 r}}$

(c) $\displaystyle \frac{2 q}{ r}\sqrt{\frac{1}{\pi \epsilon_0 r}}$

(d) $\displaystyle \frac{q}{ r}\sqrt{\frac{1}{\pi \epsilon_0 r}}$

Ans: (b)

Sol: When the spring extends by r, elastic potential energy of spring $\displaystyle = \frac{1}{2}K r^2$

Change in P.E. of configuration of charges $\displaystyle = \frac{1}{4 \pi \epsilon_0}(\frac{q^2}{r} – \frac{q^2}{r+r} )$

$\displaystyle = \frac{q^2}{8 \pi \epsilon_0 r}$

Hence , $\displaystyle \frac{1}{2}K r^2 = \frac{q^2}{8 \pi \epsilon_0 r}$

$\displaystyle K = \frac{q}{2 r}\sqrt{\frac{1}{\pi \epsilon_0 r}}$