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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 KumarAskIITians Faculty,B. Tech, IIT Delhi.
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