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The charge is on the "perpendicular bisector" of the dipole - that means the diagram looks something like this: Can you see it now?


Surely, when you consider remote point charge $e$ circling on the ring it emits radiation. That means it creates fields $\mathbf{E}$ and $\mathbf{H}$ such that they decade as $1/R$. That field is discribed by Lienard--Wiechert potentials $$ \mathbf{E} = \frac{e}{c^2R} \frac{[\mathbf{n}\times[(\mathbf{n}-\mathbf{v}/c) \times ...

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