(NOT a physics question) Is electric field always asymptopic to $x^{\alpha}$ for some rational $\alpha$?

Question

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 $R_x?$

Answer 1

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:

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

Answer 2

This is a partial answer, but maybe it'll inspire someone. Intuitively, the below considerations lead me to think there always exists an adequate $\alpha$ value.

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I'll denote the dot product with $\langle\ .\ ,\ .\ \rangle$. $$ \left(\left\| \vec{F(R_x)} \right\| x^\alpha\right)^2 = \left\langle \vec{F(R_x)},\vec{F(R_x)}\right\rangle x^{2\alpha} = \sum_{i=1}^n\sum_{j=1}^n\frac{q_iq_j}{\left\|P_i-R_x\right\|^2 \left\|P_j-R_x\right\|^2} \left\langle \vec{r_i^x},\vec{r_j^x} \right\rangle x^{2\alpha} $$ Now to simplify this expression a bit, let's assume that your ray $\vec\ell$ is the positive $x$-axis. If it isn't, just rotate the whole space, or equivalently build a coordinate system with $\vec\ell$ as the positive $x$-axis. Let $(x_i,y_i)$ be the Cartesian coordinates of point $P_i$, and $(\rho_i,\theta_i)$ its polar coordinates. For any $i$, you have $\left\| P_i-R_x \right\|^2 = \rho_i^2 + x^2 -2x\rho_i\cos\theta_i = x^2 -2xx_i+\rho_i^2$. Additionally, let $\varphi_{ij}^x$ be the angle between $\vec{r_i^x}$ and $\vec{r_j^x}$. You can express that angle explicitly in terms of $x$, $\rho_i$, $\theta_i$, $\rho_j$ and $\theta_j$, but this won't matter much here, since the angle goes to $0$ as $x$ goes to infinity. Anyway you then obtain: $$ \left(\left\| \vec{F(R_x)} \right\| x^\alpha\right)^2 =\sum_{i,j=1}^n\frac{x^{2\alpha}q_iq_j\cos\varphi_{ij}^x }{x^4-2x^3(x_i+x_j)+x^2(\rho_i^2+\rho_j^2+4x_ix_j)-2x(\rho_ix_j+\rho_jx_i)+\rho_i^2\rho_j^2} $$ From there you can use something that I think English people call a Taylor expansion (not English, so not sure), and you obtain a polynomial-like expression that is pretty easy to manipulate. I never liked this as a student and I still don't like this now, so I'll just do a second order expansion. Specifically, we know that $$ \frac{1}{1+X} = 1-X +X^2 +o(X^2) $$ so \begin{align*} &\left( \left\| \vec{F(R_x)} \right\| x^\alpha \right)^2\\ &=\sum_{i,j=1}^n \frac{x^{2\alpha}q_iq_j\cos\varphi_{ij}^x} {x^4\left( 1-2\frac{x_i+x_j}{x}+\frac{\rho_i^2+\rho_j^2+4x_ix_j}{x^2} +o\left(\frac{1}{x^2}\right) \right)}\\ &=\sum_{i,j=1}^n \frac{x^{2\alpha}q_iq_j\cos\varphi_{ij}^x}{x^4}\\ &\qquad\times \Bigg\{1- \left(-2\frac{x_i+x_j}{x}+\frac{\rho_i^2+\rho_j^2+4x_ix_j}{x^2}\right)\\ &\qquad\qquad+\left(-2\frac{x_i+x_j}{x}+\frac{\rho_i^2+\rho_j^2+4x_ix_j}{x^2}\right)^2 +o\left(\frac{1}{x^2}\right) \Bigg\}\\ &=\sum_{i,j=1}^n \frac{x^{2\alpha}q_iq_j\cos\varphi_{ij}^x}{x^4} \left( 1+2\frac{x_i+x_j}{x}-\frac{\rho_i^2+\rho_j^2+4x_ix_j}{x^2} +4\frac{\left( x_i+x_j\right)^2}{x^2} +o\left(\frac{1}{x^2}\right) \right)\\ &=\left( \sum_{i,j=1}^n q_iq_j\cos\varphi_{ij}^x \right)x^{2\alpha-4} +2\left( \sum_{i,j=1}^n q_iq_j\cos\varphi_{ij}^x\left(x_i+x_j\right) \right)x^{2\alpha-5}\\ &\qquad+\left( \sum_{i,j=1}^n q_iq_j\cos\varphi_{ij}^x\left(3x_i^2-y_i^2+3x_j^2-y_j^2+4x_ix_j\right) \right)x^{2\alpha-6} +o\left(x^{2\alpha-6}\right) \end{align*} Assuming you can treat the above expression as a regular polynomial in $x$ (which normally you shouldn't, since the $\varphi_{ij}^x$ depends on $x$), you can easily study the behaviour of this expression as $x$ goes to $+\infty$.

When $\sum_{i,j=1}^nq_iq_j=\left(\sum_{i=1}^nq_i\right)^2$ is non-zero, the polynomial is dominated by the $x^{2\alpha-4}$ term, so you must have $\alpha=2$ to have convergence. And since $\sum_{i=1}^nq_i$ is non-zero, the limit will also be non-zero. This is the case covered by The Riddler's answer.

When the sum of charges is null, you have to look at the next term $x^{2\alpha-5}\sum_{i,j=1}^nq_iq_j(x_i+x_j)$. If $\sum_{i,j=1}^nq_iq_j(x_i+x_j)$ is non-zero, you need $\alpha=5/2$. If it is zero, you have to look at the next term, etc.

If you do a large enough expansion, you can in theory find more and more conditions that your charge distribution have to respect, and that forces you to look at more and more terms in the expansion. In all likelihood, there will come a point where it is impossible for your points to simultaneously satisfy every one of these conditions, which means you can always find some value for $\alpha$ so that the limit converges and is non-zero. The limit would be equal to the coefficient in front of the leading term of the polynomial, which would have to be non-zero, and $\alpha$ would indeed be rational.

If you're unlucky, the problem structure is such that there always exist a configuration of charges/points for which every term will be zero. In that case, there are no value of $\alpha$ that satisfy your conditions.

Answer 3

The most general answer I can give you goes as follows:

The electric potential at a point $\vec{R}$ due to $n$ point charges, such that $||\vec{R}|| \ge r_{max}$, where $r_{max}$ is the furthest that any point charge is located away from the origin, is given by Wikipedia as follows:

https://en.wikipedia.org/wiki/Multipole_expansion#Spherical_form

This is called a multipole expansion of the electric potential. Essentially, the electric potential is approximated as the combination of a monopole contribution, a dipole contribution, a quadrupole contribution, an octupole contribution, and so on. The formula has the following form:

$$V(\vec{R})=\sum_{i=1}^{\infty}\frac{C_{2^{i-1}}}{r^i}=\frac{C_1}{r}+\frac{C_2}{r^2}+\frac{C_4}{r^3}+\frac{C_8}{r^4}+...$$

where $C_1$ is some constant representing the monopole contribution, $C_2$ is some constant representing the dipole contribution, $C_4$ is some constant representing the quadrupole contribution, $C_8$ is some constant representing the octupole contribution, and so on. Now strictly speaking these are not constants because they depend on the angular position of the point $\vec{R}$ in spherical coordinates (in terms of $\theta$ and $\phi$). However, since in your problem $\vec{R}$ is restricted to lie on a single ray it follows that $\theta$ and $\phi$ are constants, so $C_1,C_2,C_4,C_8,...$ are all constants as well.

Therefore, the three dimensional problem gets reduced to a one dimensional problem, with $r$ as the only independent variable. Therefore, $V(\vec{R})=V(r)$. Now crucially, in the limit as $r$ goes to $\infty$, the function $V(r)$ reduces to the largest power of $r$ with a nonzero coefficient. For example, if $C_1$ is nonzero then, in the limit, $V(r)={C_1}/{r}$.

Now I will state, without proof, the following principle: if you want $C_{2^{i-1}}$ to be zero then you need $n\ge2^i$. For example, to get rid of the monopole contribution $C_1$ you need at least $2$ charges, to get rid of the dipole contribution $C_2$ you need at least $4$ charges, to get rid of the quadrupole contribution $C_4$ you need at least $8$ charges, and so on. Put another way, if $n<2^i$ then $C_{2^{i-1}}$ is guaranteed to be nonzero.

Suppose that $\beta$ is the smallest positive integer which satisfies the condition $\beta>\log_2n$. Then, by the argument above, $V(r)$ is at least of order $1/r^{\beta}$. More specifically, $V(r)$ may be of order $1/r$ or $1/r^2$ or $1/r^3$, all the way down to $1/r^{\beta}$. Since $||\vec{F}||=\frac{d}{dr} V(r)$ it follows that $||\vec{F}||$ may be of order $1/r^2$ or $1/r^3$ or $1/r^4$, all the way down to $1/r^{\beta+1}$.

It follows that $\alpha$ can equal either $2$ or $3$ or $4$, all the way up to $\beta+1$, which is a finite positive integer. So for any finite collection of $n$ point charges located at distinct positions the answer to your question is yes, there always exists a rational $\alpha$ such that $\lim_{x\to \infty}||\vec{F}||r^{\alpha}$ is the nonzero constant $C_{2^{\alpha-2}}$.

Answer 4

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.