= Solution
Write $a=0.020\,\mathrm m$ for the side length. In an <equilateral triangle>, the distance from centre to a vertex is $a/\sqrt3$; it is two thirds of the altitude $\sqrt3a/2$. By the <superposition principle>, the <electric potential> at the centre due to the three fixed <Electrons> is
$$
\Phi(0)=-\frac{3e}{4\pi\varepsilon_0(a/\sqrt3)}.
$$
Here $e>0$ is the <elementary charge> and $\varepsilon_0$ the <vacuum permittivity>. The approaching <Electron> has <electric charge> $-e$, so its <potential energy>, measured relative to infinity, is positive:
$$
U(0)=(-e)\Phi(0)=\frac{3\sqrt3e^2}{4\pi\varepsilon_0a}.
$$
The fixed sources do not move, and radiation is excluded. Hence <conservation of mechanical energy> gives
$$
\frac12m_ev_\infty^2=U(0),\qquad\boxed{v_\infty=\sqrt{\frac{6\sqrt3e^2}{4\pi\varepsilon_0m_ea}}.}
$$
Using $e=1.60218\times10^{-19}\,\mathrm C$, $m_e=9.10938\times10^{-31}\,\mathrm{kg}$ for the <electron mass>, and $(4\pi\varepsilon_0)^{-1}=8.98755\times10^9\,\mathrm{N\,m^2\,C^{-2}}$ gives
$$
U(0)=5.99\times10^{-26}\,\mathrm J=3.74\times10^{-7}\,\mathrm{eV},\qquad\boxed{v_\infty\simeq3.63\times10^2\,\mathrm{m\,s^{-1}}.}
$$
This <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.
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