Q. Two charged particles having charges 1μC and -1μC and of mass 50g each are held at rest while their separation is 2m. Now the charges are released. Find the speed of the particles when their separation is 1m.

(a) 1/5 ms^{-1}

(b) 3/5 ms^{-1}

(c) 3/10 ms^{-1}

(d) 2/7 ms^{-1}

**Click to See Answer : **

$ \displaystyle \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r_1} + 0 = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r_2} + 2 (\frac{1}{2}m v^2) $

$ \displaystyle \frac{q_1 q_2}{4\pi \epsilon_0}(\frac{1}{r_1}-\frac{1}{r_2}) = m v^2 $

$ \displaystyle -9\times 10^9 \times 10^{-12} (\frac{1}{2}-\frac{1}{1}) = m v^2 $

$ \displaystyle m v^2 = \frac{9\times 10^{-3}}{2} $

$ \displaystyle v^2 = \frac{9\times 10^{-3}}{2\times 50 \times 10^{-3}} $

$ \displaystyle v^2 = \frac{9}{100} $

$ \displaystyle v = \frac{3}{10} m/s $