I'm struggling with one of the problems from mathematical analysis II course.
The problem is next:
Find the potential $\Phi$ and the field $\vec{\mathbf{F}} = \nabla \Phi$ for a two-dimensional dipole at the origin. with axis in the $z$ direction and dipole moment $μ$
Unfortunately I'm unable to evaluate it at all and I would love to see the solution and explanation about what is happening here and in the solution.
Thanks to everyone in advance.
The field of a dipole is essentially taking the limit of the field of two oppositely charged particles separated by a distance $\vec d$ as they come closer but in a way where the product $q \vec d$ converges to the dipole moment $\vec \mu$. So it is taking the limit of the potential field but in a very particular and specific way. The context I use is electrostatics but the same principle applies to any potential field, where $q$ would be point sources.
Specifically, given the potential field of a positively-charged particle centered at the origin to be $q\phi(\vec r)$, where $q$ is the charge and $\phi$ is field per unit charge (for convenience, will be clear below), the field generated by two particles (one positive, one negative) located at opposite sides of the origin is just $$\varphi (\vec r) = q\phi(\vec r-\frac{1}{2}\vec d) - q\phi(\vec r +\frac{1}{2} \vec d) $$ (since the field of a negative charge centered at origin is just $-q\phi(\vec r)$). Taylor expansion to first order $$\phi(\vec r-\frac{1}{2}\vec d) = \phi(\vec r)- \frac{1}{2} \nabla \phi(\vec r) \cdot \vec d + \mathcal{O}(|\vec d|^2)$$ $$\phi(\vec r+\frac{1}{2}\vec d) = \phi(\vec r)+ \frac{1}{2} \nabla \phi(\vec r) \cdot \vec d + \mathcal{O}(|\vec d|^2)$$ Combining we get $$\varphi(\vec r)=-\nabla \phi(\vec r) \cdot q\vec d + q \:\mathcal{O}(|\vec d|^2)$$ I mentioned that a dipole is just the field of two oppositely charged particles as they are brought together and $q \vec d $ converges to $\vec \mu$ in the limit, i.e. the charge $q$ is increasing to infinity while the distance $\vec d$ is decreasing to zero but in such a way that their product converges to $\vec \mu$. This means that as the charges are brought together ($\vec d \rightarrow 0$ while $q \vec d \rightarrow \vec \mu$) the field of a dipole becomes $$\varphi_{d}(\vec r)=\nabla \phi (\vec r) \cdot \vec \mu$$ since $$\lim_{q \vec d \rightarrow \vec \mu \\ \vec d \rightarrow 0} q \: \mathcal{O}(|\vec d|^2) = 0$$ So all you have to do is evaluate the gradient $\nabla \phi$ and insert it in the formula above. In electrostatics, for a charge in 3D, $\phi$ is proportional to $\frac{1}{|\vec r|}$ while for 2D, $\phi$ is proportional to $\ln(|\vec r|)$, so the gradients will be proportional to $\frac{\vec r}{|\vec r|^3}$ in 3D and $\frac{\vec r}{|\vec r|^2}$ in 2D. Your application might be different (so $\phi $ might be different) but the procedure is essentially the same.
Lastly, if you're assuming the dipole is orientation along the z-direction, then you're assuming $\vec d = d \: \hat{z}$ where $\hat{z}$ is a basis vector pointing in the z-direction, and then the moment is given by $\vec \mu = \mu \hat{z}$ so substituting into the formula, you get $$\varphi_d(\vec r) = \frac{\partial \phi}{\partial z} \mu$$