Two particles have equal masses of 0.5g each and opposite charges of +4.0*10^-5 C and -4.0*10^-5 C they are released from rest with a separation of 1.0m between them. Find the speeds of the particles when the separation is reduced to 50cm

At any position, the force acting between them,

F = -kq^2/r^2

Acceleration, a = F/m

or, a = -(kq^2/m)*(1/r^2)

or, (v dv) = -(kq^2/m)*(dr/r^2)

On intergaring both the sides with limits as when r = 1m, v = 0 and when r = 0.5m, v = v (to be calculated)

You will get v = q*root(2k/m)

You can put the values and get the answer.

Thanks and Regards,
Deepak Kumar
AskIITians Faculty,
B. Tech, IIT Delhi.