(NOT a physics question) Is electric field always asymptopic to $x^{\alpha}$ for some rational $\alpha$? In three dimensional space with origin $O$, you pick a finite number of points $P_1, P_2, \cdots, P_n$. To each point $P_i$ you assign a nonzero integer (positive or negative) $q_i$. For all other points $R$ in the plane, define the vector valued function $$\displaystyle \vec{F(R)} = \sum_{i = 1}^{n} \frac{q_i}{D(P_i, R)^2} \vec{r_i}, $$ where $D(P_i, R)$ is the Euclidean distance between $P_i, R$, and $r_i$ is a vector of unit magnitude directed from $P_i$ to $R$. Now you pick a ray $\vec{\ell}$ originating from $O$ in any direction. 

Is it true that for any such configuration of such points, there always exist an rational number $\alpha$ such that $\displaystyle \lim_{x \rightarrow \infty} \| F(R_x) \| x^{\alpha}$ converges to some nonzero constant, where $R_x \in \ell$  with $D(O, R_x) = x$ and $\| F(R_x) \|$ is the magnitude of the function at $R_x?$
No. Place positive unit charges at $(\pm 1,0,0)$ and negative unit charges at $(0,\pm 1,0)$. Then the field is zero on the whole $z$-axis.
Far away from any localized distribution of point charges the value of $||F(R_x)||$ will drop off as $1/x^2$ so you want $\alpha=2.$ In particular:

$$\lim_{x\to \infty}||F(R_x)||x^2=\sum_{i=1}^nq_i$$