= Solution
The <electric dipole moment> is $\mathbf p=qa\,\hat{\mathbf y}$. Write $\theta$ for the angle measured from $+y$, with $\hat{\boldsymbol\theta}$ pointing toward increasing $\theta$. The exact <electric potential> is
$$
\phi(\mathbf r)=k_eq\left(\frac1{|\mathbf r-(a/2)\hat{\mathbf y}|}-\frac1{|\mathbf r+(a/2)\hat{\mathbf y}|}\right).
$$
For $r\gg a$, expand the reciprocals to first order in $a/r$. Their common $1/r$ terms cancel, giving the <electric dipole> approximation
$$
\boxed{\phi(r,\theta)=\frac{k_ep\cos\theta}{r^2},\qquad
\mathbf E=\frac{k_ep}{r^3}\left(2\cos\theta\,\hat{\mathbf r}+\sin\theta\,\hat{\boldsymbol\theta}\right).}
$$
The radial <electric field> follows from $E_r=-\partial\phi/\partial r$, and the tangential <electric field> from $E_\theta=-(1/r)\partial\phi/\partial\theta$. The omitted terms are of relative order $a^2/r^2$ away from zeros of a component. Multiplying by the test <electric charge> gives the <force> components
$$
\boxed{F_r=\frac{2k_epQ\cos\theta}{r^3},\qquad F_\theta=\frac{k_epQ\sin\theta}{r^3}.}
$$
The <electric field lines> emerge from the positive <electric charge> and terminate on the negative <electric charge>. They are mirror-symmetric about the <electric dipole> axis and bow out across its equatorial plane. In the point-<electric dipole> limit their meridional curves obey $dr/(r\,d\theta)=E_r/E_\theta=2\cot\theta$, hence $r=C\sin^2\theta$.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019-dipole-field-and-charge-path.png]
{title=Electric dipole field lines and the released charge's ideal trajectory}
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