 Solution

Write $a=0.020\mathrm{m}$ for the side length. In an equilateral triangle, the distance from centre to a vertex is $a/\sqrt{3}$; it is two thirds of the altitude $\sqrt{3}a/2$. By the superposition principle, the electric potential at the centre due to the three fixed Electrons isHere $e>0$ is the elementary charge and ${\epsilon }_0$ the vacuum permittivity. The approaching Electron has electric charge $-e$, so its potential energy, measured relative to infinity, is positive:The fixed sources do not move, and radiation is excluded. Hence conservation of mechanical energy givesUsing $e=1.60218\times 10^{-19}\mathrm{C}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ for the electron mass, and $(4\pi {\epsilon }_0{)}^{-1}=8.98755\times 10^9\mathrm{N}{\mathrm{m}}^2{\mathrm{C}}^{-2}$ givesThis speed is small enough to justify classical kinetic energy. There is an idealization in the word “stops”: along the line normal to the triangle through its centre, this threshold energy makes the Electron approach the central equilibrium asymptotically. A smooth, conservative trajectory cannot reach an equilibrium with exactly zero velocity in finite time unless it was already there. The energy threshold, and therefore the requested initial speed, is nevertheless unambiguous.