= Solution
In a <Cartesian coordinate system> the leading <electric dipole field> gives
$$
\mathbf F(x,y)=\frac{k_epQ}{(x^2+y^2)^{5/2}}\left(3xy\,\hat{\mathbf x}+(2y^2-x^2)\,\hat{\mathbf y}\right).
$$
Therefore
$$
\boxed{\mathbf F(-d,0)=\mathbf F(d,0)=-\frac{k_epQ}{d^3}\hat{\mathbf y},\qquad
\mathbf F(0,-d)=\frac{2k_epQ}{d^3}\hat{\mathbf y}.}
$$
The <force> is downwards on the equator and upwards on the negative axial side. A free positive <electric charge> released from rest at the left-hand point initially accelerates downwards. As it moves into $x<0,y<0$, the <force> also acquires a positive $x$ component. Its path bends toward the right; the <force> continually changes as its position changes. An initial velocity was not specified, so the following more precise trajectory assumes release from rest and neglects <radiation reaction>.
There is a useful exact result within the point-<electric dipole> approximation. Put $A=k_epQ>0$; the <potential energy> is $U=Ay/r^3$, a <homogeneous function> of degree $-2$. Thus $\mathbf r\cdot\mathbf F=2U$. For a particle of mass $m$, <Newton's second law> and <conservation of energy> imply
$$
\frac{d^2r^2}{dt^2}=2v^2+\frac2m\mathbf r\cdot\mathbf F=\frac4m(K+U)=\frac{4E}{m}.
$$
At release, both <kinetic energy> and <potential energy> vanish, so $E=0$; also $d(r^2)/dt=0$. Consequently $r=d$ throughout the motion. Parametrize the lower semicircle by $x=-d\cos\alpha$, $y=-d\sin\alpha$. Then
$$
\dot\alpha^2=\frac{2A}{md^4}\sin\alpha,\qquad
\ddot\alpha=\frac{A}{md^4}\cos\alpha.
$$
The <electric charge> gains speed until the bottom point, where the <force> is inward and the speed is $\sqrt{2A/(md^2)}$. It then slows, reaches $(d,0)$ at rest, and reverses. Thus, under these assumptions,
$$
\boxed{\text{the charge oscillates along the lower semicircle }x^2+y^2=d^2,\quad y\leq0.}
$$
Its trajectory is not an <electric field line>: the <force> determines acceleration, rather than velocity direction. For the finite two-charge <electric dipole>, corrections of order $a^2/d^2$ perturb the circular trajectory; an arbitrary nonzero initial velocity also changes it.
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