Zero potential locus of two charges in 2D plane

Template:Nocat Suppose we place two charges $q_1$ and $q_2$ in the cartesian plane in the points $A(x_1,0)$ and $B(x_2,0)$ and we want to find the locus where the potential $V$ of the field is zero (Notice that in order to simplify the problem we put the two charges lying in the x axis).

Without any loss of generality we can assume that $q_1=\lambda q_2$ (1), where λ is a costant. In order for the net potential of the field to give zero, the two charges need to be of opposite signs and thus we conclude that λ in equation (1) is a negative constant. Consequently, it is true that $\lambda \in (-\mathrm{\infty },0)$(2).

Now, lets assume there is a point

$M(x,y)$

in 2D space which satisfies the condition that its potential is zero. From the superposition principle it is true that:

$$V_{(M)}=V_{(A)\to (M)}+V_{(B)\to (M)}=0$$

which expands to:

$$V_{(M)}=K\frac{q_1}{(AM)}+K\frac{q_2}{(BM)}=0$$

.From the assumption (1) that we introduced, the equation above transforms to:

$$V_{(M)}=K(\frac{q_1}{(AM)}+\frac{q_2}{(BM)})=K{q}_1(\frac{1}{(AM)}-\frac{|\lambda |}{(BM)})=0(3)$$$K$

and

$q_1$

are non-zero constants so equation (3) becomes:

$$(BM)=|\lambda |(AM)(4)$$

Potential locus of two charges

It is true that:

$$(AM)=\sqrt{(x-x_1{)}^2+y^2}and(BM)=\sqrt{(x-x_2{)}^2+y^2}$$and (4) transforms to:$$\sqrt{(x-x_2{)}^2+y^2}=|\lambda |\sqrt{(x-x_1{)}^2+y^2}$$Now both sides of the equation above are positive and thus we can square both sides:$$(x-x_2{)}^2+y^2={\lambda }^2[(x-x_1{)}^2+y^2]\iff (1-{\lambda }^2)x^2+(1-{\lambda }^2)y^2+2x({\lambda }^2{x}_1-x_2)+x_2^2-{\lambda }^2{x}_1^2=0(5)$$If $\lambda \in (-\mathrm{\infty },-1)\cup (-1,0)$ (5) becomes:$$x^2+y^2+2(\frac{x_2-{\lambda }^2{x}_1}{1-{\lambda }^2})x+(\frac{x_2^2-{\lambda }^2{x}_1^2}{1-{\lambda }^2})=0(6)$$Equation (6) describes a circle (in the form $x^2+y^2+Ax+C=0$) with its centre moving in the x' axis.

Its centre K is$$K\equiv (-A/2,0)\equiv (\frac{{\lambda }^2{x}_1-x_2}{{\lambda }^2-1},0)$$and its radius ($r=\frac{\sqrt{A^2-4C}}{2}$):$$r=|\frac{x_2-x_1}{\lambda +{\lambda }^{-1}}|=\frac{d(A,B)}{|\lambda +{\lambda }^{-1}|}$$Now we need to check the case of $\lambda =-1$:

From the equation (5) we get:$$2x(x_2-x_1)+x_2^2-x_1^2=0\iff 2x(x_1-x_2)+(x_2-x_1)(x_1+x_2)=0\iff x=\frac{x_1+x_2}{2}$$So the zero potential locus is either a circle when $\lambda \ne -1$ or the perpendicular bisector of (AB) when $\lambda =-1$ and thus $(q_1=-q_2)$.