Solution: 2D dipole with orientation angle $\alpha$.

We derive the expression for a 2D dipole from the superposition of a source and a sink with the same strength $m$ and located a distance $2a$ appart from each other. The superposition of the source and sink follows:

$$\phi(x,y) = \frac{m}{2\pi }\left[ \ln\sqrt {(x - a\cos\alpha)... ...\sqrt {(x + a\cos\alpha)^2 + (y+a\sin\alpha)^2} \right] \notag$$

We take the limit $a \rightarrow 0$. To preserve a finite effect from the two singularities as they are brought togheter, their strength $m$ must increase at the same time, for otherwise they will cancel out in the limit when they coincide. Thus it is necessary to make the product $\mu = 2ma$ a constant, with the result

$$\begin{align}\phi(x,y) = & \lim_{a \to 0} \frac{\mu}{2\pi a}\left[ \ln\sqrt {(x ... ...frac{\mu}{2\pi}\frac{x\cos \alpha + y\sin \alpha }{x^2 + y^2} \notag \end{align}$$

NOTE: dipole $= \mu \frac{\partial }{\partial \xi }$ (unit source)