= Solution
Write $k=1/(4\pi\epsilon_0)$. The separation of the two <point charges> is $2l\sin\phi$, so <Coulomb's law> gives the horizontal repulsive <force> $F=kQ^2/(4l^2\sin^2\phi)$. Resolving the rod's <tension> vertically and horizontally gives $T\cos\phi=mg$ and $T\sin\phi=F$. Hence
$$
\boxed{\sin^2\phi\tan\phi=\frac{Q^2}{16\pi\epsilon_0mgl^2}.}
$$
The left side increases from zero to infinity for $0<\phi<\pi/2$, so this determines a unique outward <equilibrium>. For a small angle, $\phi^3\simeq Q^2/(16\pi\epsilon_0mgl^2)$. With $g=9.81\,\mathrm{m\,s^{-2}}$, direct numerical solution gives
$$
\boxed{\phi=7.71\times10^{-3}\,\mathrm{rad}=0.442^\circ.}
$$
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