= Solution
A large <electrical conductor> between the <point charges> provides <electrostatic shielding>: one must not retain their original mutual <Coulomb force> through an ideal infinite plate. In the usual broad-plate approximation, let $\sigma$ denote the unperturbed <surface charge density> \b[on each face]. The <electrostatic boundary condition> gives an outward <electric field> $E_0=\sigma/\epsilon_0$. Neglecting local <electrostatic induction>, the rod's <tension> then satisfies $mg\tan\phi=QE_0$, giving
$$
\boxed{\phi\simeq\tan^{-1}\!\left(\frac{Q\sigma}{\epsilon_0mg}\right)=1.98^\circ.}
$$
There are two physical qualifications. First, if $\sigma$ means the \b[total charge per projected plate area], shared equally between the two faces, then $E_0=\sigma/(2\epsilon_0)$ and the same approximation gives $\phi\simeq0.989^\circ$. The factor of two is a convention about the specified <surface charge density>, not a different <Gauss's law>.
Second, a <point charge> near an <electrical conductor> makes its <surface charge density> nonuniform. For an infinite plane with prescribed background <electric field> $E_0$, the <method of images> adds an attractive <force> $kQ^2/(4x^2)$, where $x=l\sin\phi$. The more accurate <equilibrium> equation is
$$
mg\tan\phi=QE_0-\frac{Q^2}{16\pi\epsilon_0l^2\sin^2\phi}.
$$
Its larger, stable root is approximately $1.96^\circ$ for the per-face convention, or $0.877^\circ$ for the total-area convention. This ideal-plane refinement is not an exact answer for an unspecified finite plate: its dimensions and the meaning of the quoted <surface charge density> would be needed. In particular, adding a uniform-sheet <electric field> to the original mutual repulsion would describe a different, nonshielding model.
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