= Solution
Here $q_1=3.8q_3$ and $q_2=-1.9q_3$. The part of the <electrostatic potential energy> that changes is
$$
U(f)=\frac{k_eq_3^2}{L}\left(\frac{3.8}{1+f}-\frac{1.9}{1-f}\right).
$$
At the two endpoints, $U(-0.9)=37k_eq_3^2/L$ and $U(0.9)=-17k_eq_3^2/L$. The fixed-source interaction energy is unchanged, and conservation of energy gives
$$
\frac12mv^2=U(-0.9)-U(0.9)=\frac{54k_eq_3^2}{L},\qquad
\boxed{v=\sqrt{\frac{108k_eq_3^2}{mL}}.}
$$
The force points toward increasing $f$ throughout this interval: the left charge repels $q_3$ and the right charge attracts it, so release from rest reaches the stated endpoint without a turning point.
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