# How do I find the neutral point in an electric field created by three or more charges?

I know that for two charges $$q_1$$ and $$q_2$$ of the same polarity, the neutral point will be at the internal section of the segment by the ratio of $$\sqrt q_1$$:$$\sqrt q_2$$, and if they are of opposite polarity the point will be at the external division by the same ratio.

Now, for 3 charges I tried to compare this with the center of mass formula. For two mass $$m_1$$ and $$m_2$$ the center of mass will be found at the internal section with the ratio of $$\frac{1}{m_1}:\frac{1}{m_2}$$

So I came to the formula of point (x,y) being the neutral point : $$(x,y) = (\frac{\sum_{i=0}^n\frac{x_i}{\sqrt q_i}}{\sum_{i=0}^n\frac{1}{\sqrt q_i}}, \frac{\sum_{i=0}^n\frac{y_i}{\sqrt q_i}}{\sum_{i=0}^n\frac{1}{\sqrt q_i}})$$

But it did not work. I did not find any good resource for finding the neutral point for three or more charges. So here I am.

• by neutral point do you mean the point in space where the net force due to all charges is zero? – lineage Jan 31 at 13:44
• – Frobenius Feb 1 at 15:59
• @lineage I agree. All points that are not charges are neutral. – JEB Feb 1 at 18:51

There is in general no simple closed formula for the positions of the neutral points of a system of 3 or more charges. The centre-of-mass formula does not apply because the neutral points have no connection with the centre of mass. There can be several neutral points but only one centre of mass.

If charges $$Q_i$$ are placed at points ($$x_i, y_i, z_i$$) then the resultant electric field at a point P ($$x_0, y_0, z_0$$) is zero when the sum of electric field components at P in the $$x, y$$ and $$z$$ directions are separately zero : $$\sum \frac{Q_i X_i}{R_i^3}=\sum\frac{Q_i Y_i}{R_i^3}=\sum\frac{Q_i Z_i}{R_i^3}=0$$ where $$X_i=x_i-x_0, Y_i=y_i-y_0, Z_i=z_i-z_0, R_i^2=X_i^2+Y_i^2+Z_i^2$$.

In general the above triplet of equations will be coupled and transcendental, having only numerical solutions. For arrangements which have some symmetry it may be possible to simplify the calculation by pairing charges about a single line of symmetry. This will eliminate two co-ordinates, reducing the system to a single-variable polynomial equation of degree 8.

For example, suppose identical charges $$Q$$ are placed at the vertices AB and charges $$2Q$$ at vertices CD of a rectangle ABCD of dimensions $$3\times 2$$. Consider the electric field at point P which lies on the perpendicular bisector of AB and CD, as in the following diagram : The resultant field $$\mathbf{E_1}=2kQ\frac{x}{r^3}$$ due to charges AB and that $$\mathbf{E_2}=2k(2Q)\frac{y}{s^3}$$ due to charges CD point in opposite directions in the same straight line. Here $$y=3-x, r^2=1+x^2, s^2=1+y^2$$. The resultant electric field at P will be zero when $$\frac{x}{(1+x^2)^{3/2}}=\frac{2y}{(1+y^2)^{3/2}}$$

This must be solved numerically. Wolfram Alpha gives the possible roots $$x \approx 0.236168, 0.990072, 2.95105$$.

A further 2 solutions would be found from considering off-axis points. This is most easily seen from symmetry for 4 charges arranged in a square, then deforming the square.

The following diagram illustrates the variation of the electric fields $$E_1, E_2$$ due to charges AB, CD respectively when the charges are the same on each pair. The curves intersect at 3 points, which are the null points. For $$n$$ charges arranged in a (possibly irregular) convex polygon there will be $$n+1$$ neutral points all lying within the polygon. [reference or proof required]

All neutral points are points of unstable equilibrium. This can be seen by applying Gauss' Law to a small volume surrounding the neutral point. The Gaussian surface contains no charge so the total flux across its surface is zero; it must have as many lines of flux leaving as there are entering. Stable points must have all field lines converging on them from every direction. This is incompatible with the total flux being zero.