Closest Approach

1) By Newton's 3rd law, the forces on the two particles have identical magnitudes

2) Therefore, the magnitudes of the impulses on the particles are the same

3) Since the masses are the same, the magnitudes of the changes in velocity are the same

4) There is therefore some $\Delta v$ which both particles will have experienced when they are closest together. At this point, the maximum amount of the initial kinetic energy will have been converted into potential energy. So we are looking for the $\Delta v$ which minimizes the kinetic energy.

5) KE equation: $KE = \frac{1}{2} m (v_0-\Delta v)^{2} + \frac{1}{2} m (\Delta v)^{2}$

6) Differentiating with respect to $\Delta v$ and setting to zero: $-(v_0-\Delta v) + \Delta v = 0$

7) Thus, $\Delta v = \frac{v_0}{2}$ . From this, it is evident that half of the original kinetic energy is stored as potential energy. We can then use the Coulomb potential energy expression to solve for the minimum separation distance.

8) $k_e q^{2} (\frac{1}{d_{min}} - \frac{1}{1}) = \frac{1}{4}m v_0^{2}$

9) Plugging in numbers gives $d_{min} \approx 0.783 \, m$