= Solution
Interpret the coordinates as a plane in ordinary three-dimensional <electrostatics>. Set $k_e=1/(4\pi\epsilon_0)$, where $\epsilon_0$ is the <vacuum permittivity>. The work required to bring the second <point charge> in quasistatically is its <electric charge> times the first charge's <electric potential>:
$$
\boxed{U=-\frac{k_eQ^2}{d}.}
$$
The individual point-charge self-energies are excluded; this is the <electrostatic potential energy of point charges> relative to infinite separation.
By the <superposition principle>, for $\mathbf r=(x,y)$ and $\mathbf d=(d,0)$ the <electric field> is
$$
\mathbf E(\mathbf r)=k_eQ\left(\frac{\mathbf r}{|\mathbf r|^3}-\frac{\mathbf r-\mathbf d}{|\mathbf r-\mathbf d|^3}\right).
$$
At the first position both contributions point toward positive $x$, whereas at the second they oppose one another. Thus
$$
\boxed{\mathbf E(d/3,0)=\frac{45k_eQ}{4d^2}\,\hat{\mathbf x},\qquad
\mathbf E(-d/3,0)=-\frac{135k_eQ}{16d^2}\,\hat{\mathbf x}.}
$$
At $(d/2,d)$ the two distances are $\sqrt5d/2$. The vertical <electric field> components cancel and the horizontal components add:
$$
\boxed{\mathbf E(d/2,d)=\frac{8k_eQ}{5\sqrt5\,d^2}\,\hat{\mathbf x}.}
$$
\b[Dimensional convention.] The source's description of space as two-dimensional cannot literally mean two-dimensional <electrostatics> with the same inverse-square formula. In a genuinely two-dimensional <Gauss's law>, the circumference grows as $r$, giving an <electric field> proportional to $1/r$ and a logarithmic <electric potential>. For example, integrating $1/r$ from $d$ to infinity diverges, so the same finite reference <potential energy> at infinite separation does not exist. The displayed results use the conventional three-dimensional <Coulomb's law> restricted to the stated plane.
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