# Two identical particles , each having a charge of $\;2\times10^{-4}\;C\;$ and mass 10g are kept at a separation of 10 cm and released . What would be the speeds of the particles when the separation becomes large ?

$(a)\;600\;m/s\qquad(b)\;600\;cm/s\qquad(c)\;60\;m/s \qquad(d)\;None\;of\;these$

Answer : (a) $\;600\;m/s$
Explanation :
Initial P . E=$\;\large\frac{9\times10^{9}\times(2\times10^{-4})^2}{0.1}$
$=9\times10^{9}\times10^{1}\times4\times10^{-8}$
$P .E_{i} =36\times10^2\;J$
Final P . E : $P . E_{f} =0$
Using conservation of momentum and energy
$2\;\large\frac{1}{2}\;mv^2=36\times10^2$
$10^{-2}\times v^2=36\times10^2$
$v=6\times10^2\;m/s\;.$

The total energy will be conserved Ki + Ui = Kf + Uf 0 + 9×10^9 × q1q2/r = 1/2 mv^2 + 1/2 mv^2 +0 Or 9×10^9 × 2× 10^-4 × 2× 10^-4 / 0.1 = (10×10^-3) × v^2 V^2 = 3600×100 V = 600 m /s