A short electric dipole is placed near a long, straight, uniformly...

The electric field $\vec{E}$ due to the uniformly positively charged wire points radially outward (away from the wire).

The magnitude of this electric field at a distance $r$ is $E = \dfrac{\lambda}{2\pi\epsilon_0 r}$, which decreases as $r$ increases.

The dipole moment $\vec{p}$ points towards the wire. This implies that the positive charge of the dipole is closer to the wire, while the negative charge is farther away.

Since the electric field is stronger closer to the wire, the repulsive force on the positive charge is greater in magnitude than the attractive force on the negative charge.

Therefore, the net force on the dipole is directed away from the wire, causing it to move away.

The electrostatic potential energy of the dipole is given by $U = -\vec{p} \cdot \vec{E}$.

Since $\vec{p}$ points towards the wire and $\vec{E}$ points away from it, the angle between them is $180^\circ$. Thus, $U = -pE \cos(180^\circ) = pE$.

As the dipole moves away from the wire, the electric field $E$ decreases, which means its potential energy $U$ will decrease. This is also consistent with the fact that a system free to move will always move in a direction that lowers its potential energy.

Answer: Away from the wire, as its potential energy will decrease.