Poisson equation in 2D and 3D: geometrical reason for the difference The Poisson equation in 3D shows a fundamental solution in 3D which decays with $\sim 1/r$, whilst in 2D it shows a much different decay $\sim -\ln r$. While in 3D not only the solution, but also its integral decays, in 2D the integral does not. 
I have been wondering what is the geometrical (or better, physical, for example in the case of a density of electric charges) reason underlying this profound difference. 
Is it related with the fact that the stream line of any fields has (for purely geometrical reasons) less freedom to be, for instance, skew and not only intersecting or parallel to another?
 A: The best interpretation I can see of this fact is related to Gauss' theorem. Poisson equation is of the form
$$\nabla^2\phi = \rho,$$
but if you set $\mathbf F = \nabla\phi$, this is also an equation for the vector field $\mathbf F$,
$$\nabla\cdot\mathbf F = \rho.$$
Let us consider a density distribution given by a total mass $Q$ centered at a single point, say $\rho(x) = Q\delta(x)$, where $\delta$ is the Dirac's delta. By Gauss' theorem, the total flux around a closed surface $\Sigma$ centered at the origin of the axis is equal to the total "charge" in the enclosed volume. In three dimensions, the surface scales as $r^2$, where $r$ is the distance from the origin, and this compensates the $1/r^2$ from the field $\mathbf F$ (since $\phi\sim1/r$ implies $\mathbf F\sim1/r^2$. If you apply the same argument to a 2-dimensional "world", then $\phi\sim\log r$, hence $\mathbf F\sim1/r$, but now the "surface" $\Sigma$, which is just a closed curve, scales as $r$, so even in this case the flux of $\mathbf F$ is constant over every closed path enclosing the origin. This argument of course generalises to an arbitrary number of dimensions, and this is seen by the most general form of Gauss's theorem, i.e. Stokes' theorem
$$\int_{\partial\Omega}\alpha = \int_\Omega\text d\alpha.$$
A: Phoenix87 has already given a correct answer. Let us here just try to stress the main point: It's Gauss's law
$$ \Phi_E~=~ \frac{Q}{\varepsilon_0}. $$
In $d$ spatial dimensions, the electric field caused by a point charge $Q$ is therefore 
$$ |\vec{E}(r)|~=~ \frac{Q}{\varepsilon_0 ~{\rm Vol}(S^{d-1})} ~\propto~ r^{1-d}, $$
because the surface-volume of a $d-1$ sphere$^1$ with radius $r$ scales as $r^{d-1}$.
Then the electric potential becomes 
$$ \phi(r) ~\propto~ \left\{\begin{array}{rcl} r^{2-d} &\text{for}& d~\neq~ 2, \cr
\ln r &\text{for}& d~=~2. \end{array} \right. $$
--
$^1$ Not to be confused with a ball.
