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}} $