View source for Zero potential locus of two charges in 2D plane

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Suppose we place two charges <math>q_{1}</math> and <math>q_{2}</math> in the cartesian plane in the points <math>A(x_{1},0)</math> and <math>B(x_{2},0)</math> and we want to find the locus where the potential <math>V</math> 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 <math>q_{1} = \lambda q_{2}</math> '''(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 <math>\lambda\in(-\infin,0)</math>'''(2).'''

Now, lets assume there is a point <math>M(x,y)</math> in 2D space which satisfies the condition that its potential is zero. From the superposition principle it is true that:<math display="block">V_{(M)}=V_{(A)\rightarrow(M)}+V_{(B)\rightarrow(M)}=0</math>which expands to:<math display="block">V_{(M)}=K\frac{q_{1}}{(AM)}+K\frac{q_{2}}{(BM)}=0</math>.From the assumption (1) that we introduced, the equation above transforms to:<math display="block">V_{(M)}=K(\frac{q_{1}}{(AM)}+\frac{q_{2}}{(BM)})=Kq_{1}(\frac{1}{(AM)}-\frac{|\lambda|}{(BM)})=0\qquad(3)</math><math>K</math> and <math>q_{1}</math> are non-zero constants so equation '''(3)''' becomes:<math display="block">(BM)=|\lambda|(AM) \ \ (4)</math>[[File:Charges (A) and (B) and a random point M(x,y).png|center|Potential locus of two charges]]It is true that:

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

Its centre K is<math display="block">K\equiv(-A/2 , 0)\equiv(\frac{\lambda^2x_{1}-x_{2}}{\lambda^2-1},0)</math>and its radius (<math>r = \frac{\sqrt{A^2-4C}}{2}</math>):<math display="block">r=|\frac{x_{2}-x_{1}}{\lambda+\lambda^{-1}}|=\frac{d(A,B)}{|\lambda+\lambda^{-1}|}</math>Now we need to check the case of <math>\lambda = -1</math>:

From the equation '''(5)''' we get:<math display="block">2x(x_{2}-x_{1})+x_{2}^2-x_{1}^2=0\Leftrightarrow2x(x_{1}-x_{2})+(x_{2}-x_{1})(x_{1}+x_{2})=0\Leftrightarrow x=\frac{x_{1}+x_{2}}{2}</math>So the zero potential locus is either a circle when <math>\lambda\neq-1</math> or the perpendicular bisector of '''(AB)''' when <math>\lambda=-1</math> and thus <math>(q_{1}=-q_{2})</math>.