 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 on each face. The electrostatic boundary condition gives an outward electric field $E_0=\sigma /{ϵ}_0$. Neglecting local electrostatic induction, the rod's tension then satisfies $mg\tan \varphi =Q{E}_0$, givingThere are two physical qualifications. First, if $\sigma$ means the total charge per projected plate area, shared equally between the two faces, then $E_0=\sigma /(2{ϵ}_0)$ and the same approximation gives $\varphi \simeq 0.989^{\circ }$. 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 $k{Q}^2/(4x^2)$, where $x=l\sin \varphi$. The more accurate equilibrium equation isIts 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.