To solve this problem it is necessary to resort to the energy conservation equations, both kinetic and electrical.
By Coulomb's law, electrical energy is defined as
[tex]EE = \frac{kq_1q_2}{d}[/tex]
Where,
EE = Electrostatic potential energy
q= charge
d = distance between the charged particles
k = Coulomb's law constant
While kinetic energy is defined as
[tex]KE = \frac{1}{2} mv^2[/tex]
Where,
m= mass
v = velocity
There by conservation of energy we have that
EE= KE
There is not Initial kinetic energy, then
[tex]\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = 2*\frac{1}{2}mv_f^2[/tex]
[tex]\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = mv_f^2[/tex]
[tex]v_f^2= \frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} }{m}[/tex]
[tex]v_f = \sqrt{\frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'}}{m}}[/tex]
Replacing with our values we have,
[tex]v_f = \sqrt{\frac{\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}}{5.50*10^{-15}}}[/tex]
[tex]v_f = 2.802*10^7m/s[/tex]
Therefore the speed of particle B at the instat when the particles are 3m apart is [tex]2.802*10^7m/s[/tex]
