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Electric and gravitational fields have many similarities, and this question is applicable to both, but for the framing of this question, in particular, we will be using an electric field. Imagine the field defined as $\vec{E}(x,y,z) = \kappa \sum_{i=1}^{i=N} \frac{Q_i}{r_i^2} \hat{r}_{Qi}$ . To clarify the notation here, $Q_i$ is the charge of the particle in field, $r_i$ is the distance to the particle

(defined as $r = \sqrt{(x_1 -x)^2+(y_1-y)^2 + (z_2 - z)^2}$ ), and $\hat{r}_{Qi}$ is the unit vector between the point charge and the charge $Q$.

Can anyone find a mathematical description of all of the points in this field where the field strength is 0? Does the number of solutions change with the number of charges $N$? In which cases are there solutions and when are there none? An extension of the same question: At what points is $\vec{E}(x,y,z) = P $ , where $P$ is an arbitrary value of charge?

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Can anyone find a mathematical description of all of the points in this field where the field strength is $0$?

Well, I never say never, but I would be very surprised if it were possible to find a closed-form expression because the problem is so general. These $N$ charges have all different possible values of $Q_i$ and of $r_i$. More importantly, all of the $\vec{r}_i$ vectors actually depend on your chosen field point, so you can't just solve for $\vec{r}_N$ (for example) by setting the whole expression to zero and rearranging. On the other hand, I wouldn't be surprised if there were an algorithm that would let you find zero points — but it's beyond me.

Does the number of solutions change with the number of charges $N$?

Certainly. (I'll be assuming that every $Q_i$ is nonzero.) Obviously if $N=0$, there is no electric field so your solution is everywhere. For $N=1$, you can actually evaluate the field strength everywhere, find that it's $1/r$, which is never zero at finite distance, so there's no solution. The problem there is that there's no other charge that could cancel the field from that first charge. But for $N>1$, we always have the possibility of canceling charges.

In fact, I can show that there are configurations for any $N>1$ where there is a solution: you choose $N$ and decide what all the charges $Q_i$ are, and I get to arrange those charges to show that there's a configuration with a solution. To be precise, the solution will be the origin of our coordinates as long as the charges are placed at these positions: \begin{equation} \vec{r}_j = \mathrm{sgn}(Q)\, \sqrt{|Q|} \left( \hat{x}\cos\theta_j + \hat{y}\sin\theta_j \right), \qquad \text{where} \qquad \theta_j = \frac{2\pi j} {N}. \end{equation} (Note that this could, depending on how you choose the $Q_i$, result in charges that are in the same exact place. Getting them there and keeping them there are your jobs.) Of course, if you also get to decide where the charges go, I don't know that there will always be a solution.

An extension of the same question: At what points is $|\vec{E}|=P$, where $P$ is an arbitrary value of charge?

[I've added the magnitude sign to $\vec{E}$ so that the question makes a little more sense. Of course, that can't equal a charge because the units are wrong. So let's just say $P$ is something nonzero that you choose.] Take the same solution as I gave above. The electric field strength at the origin is zero. Now go directly towards any one of the charges, and the magnitude of the field will increase all the way to $\infty$ as you approach that charge. (The direction may change as you move there, but that doesn't matter.) Somewhere along that path, you will necessarily find the magnitude $P$.

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