To solve this problem it is necessary to apply the law of conservation of Energy and the law of Coulomb.
This way we have to
[tex]\Delta EE = \Delta KE[/tex]
Where,
[tex]\Delta EE[/tex] = Potential Electric Energy
[tex]\Delta KE =[/tex] Kinetic Energy
Therefore,
[tex]\frac{kq_Aq_B}{d}-\frac{kq_Aq_B}{d'}= \frac{1}{2}mv^2+\frac{1}{2}mv^2[/tex]
[tex]\frac{kq_Aq_B}{d} = \frac{kq_Aq_B}{d'}+mv^2[/tex]
Replacing we have that
[tex]\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}=\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}+5.5*10^{-15}v^2[/tex]
[tex]5.5*10^{-15}*v^2 = \frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}[/tex]
[tex]v = 2.8*10^7m/s[/tex]
Therefore the speed of particle B at the instant is [tex]2.8*10^7m/s[/tex]
