# Zero potential locus of two charges in 2D plane

Suppose we place two charges ${\displaystyle q_{1}}$ and ${\displaystyle q_{2}}$ in the cartesian plane in the points ${\displaystyle A(x_{1},0)}$ and ${\displaystyle B(x_{2},0)}$ and we want to find the locus where the potential ${\displaystyle V}$ 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 ${\displaystyle q_{1}=\lambda q_{2}}$ (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 ${\displaystyle \lambda \in (-\infty ,0)}$(2).

Now, lets assume there is a point ${\displaystyle M(x,y)}$ in 2D space which satisfies the condition that its potential is zero. From the superposition principle it is true that:

${\displaystyle V_{(M)}=V_{(A)\rightarrow (M)}+V_{(B)\rightarrow (M)}=0}$
which expands to:
${\displaystyle V_{(M)}=K{\frac {q_{1}}{(AM)}}+K{\frac {q_{2}}{(BM)}}=0}$
.From the assumption (1) that we introduced, the equation above transforms to:
${\displaystyle V_{(M)}=K({\frac {q_{1}}{(AM)}}+{\frac {q_{2}}{(BM)}})=Kq_{1}({\frac {1}{(AM)}}-{\frac {|\lambda |}{(BM)}})=0\qquad (3)}$
${\displaystyle K}$ and ${\displaystyle q_{1}}$ are non-zero constants so equation (3) becomes:
${\displaystyle (BM)=|\lambda |(AM)\ \ (4)}$

It is true that:

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

Its centre K is

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

From the equation (5) we get:

${\displaystyle 2x(x_{2}-x_{1})+x_{2}^{2}-x_{1}^{2}=0\Leftrightarrow 2x(x_{1}-x_{2})+(x_{2}-x_{1})(x_{1}+x_{2})=0\Leftrightarrow x={\frac {x_{1}+x_{2}}{2}}}$
So the zero potential locus is either a circle when ${\displaystyle \lambda \neq -1}$ or the perpendicular bisector of (AB) when ${\displaystyle \lambda =-1}$ and thus ${\displaystyle (q_{1}=-q_{2})}$.