Answer:
[tex]7.8\cdot 10^6 m/s[/tex]
Explanation:
When the electron moves, the gain in its kinetic energy is equal to the decrease in its electric potential energy, since the total energy must be conserved.
Therefore, we can write:
[tex]K_f-K_i = U_i-U_f[/tex]
where
[tex]K_f=\frac{1}{2}mv^2[/tex] is the final kinetic energy of the electron, where
[tex]m=9.11\cdot 10^{-31} kg[/tex] is its mass
v is its final speed
[tex]K_i=0[/tex] is the initial kinetic energy (the electron starts from rest)
[tex]U_i=\frac{kq_1 e}{r_1}+\frac{kq_2 e}{r_2}[/tex] is the initial electric potential energy, where
k is the Coulomb's constant
[tex]q_1=3.90\cdot 10^{-9}C[/tex] is the charge 1
[tex]e=-1.6\cdot 10^{-19}C[/tex] is the electron charge
[tex]r_1=0.23 m[/tex] is the initial distance of the electron from charge 1
[tex]q_2=1.80\cdot 10^{-9}C[/tex] is charge 2
[tex]r_2=0.23 m[/tex] is the initial distance of the electron from charge 2
[tex]U_f=\frac{kq_2 e}{r'_1}+\frac{kq_2 e}{r'_2}[/tex] is the final electric potential energy, where
[tex]r_1=0.10 m[/tex] is the final distance between electron and charge 1
[tex]r_2=0.46-0.10 = 0.36 m[/tex] is the final distance between electron and charge 2
Substituting everything we find:
[tex]U_i = \frac{(9\cdot 10^9)(3.90\cdot 10^{-9})(-1.6\cdot 10^{-19})}{0.23}+\frac{(9\cdot 10^9)(1.80\cdot 10^{-9})(-1.6\cdot 10^{-19})}{0.23}=-3.57\cdot 10^{-17} J[/tex]
[tex]U_f = \frac{(9\cdot 10^9)(3.90\cdot 10^{-9})(-1.6\cdot 10^{-19})}{0.10}+\frac{(9\cdot 10^9)(1.80\cdot 10^{-9})(-1.6\cdot 10^{-19})}{0.36}=-6.34\cdot 10^{-17} J[/tex]
So the final kinetic energy is
[tex]K_f=U_i-U_f=-3.57\cdot 10^{-17}-(-6.34\cdot 10^{-17})=2.77\cdot 10^{-17} J[/tex]
And therefore, the final speed is:
[tex]v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(2.77\cdot 10^{-17})}{9.11\cdot 10^{-31}}}=7.8\cdot 10^6 m/s[/tex]
