At a distance a apart, at points A and B , there are two…

At a distance \(a\) apart, at points \(A\) and \(B\), there are two oppositely charged electric charges \(+q\) and \(-q\). Taking point \(A\) as the coordinate origin and the line \(AB\) as the \(x\)-axis, formulate the equation of the equipotential lines of the electric field created by these charges.

1. Potential of each point charge and superposition

With \(A=(0,0)\) carrying \(+q\) and \(B=(a,0)\) carrying \(-q\), and \(r_A=\sqrt{x^2+y^2}\), \(r_B=\sqrt{(x-a)^2+y^2}\), the total potential at \(P(x,y)\) is \[ V(x,y)=kq\left(\frac{1}{r_A}-\frac{1}{r_B}\right). \] An equipotential line is the locus \(V(x,y)=V_0\), i.e., writing \(C=V_0/(kq)\), \[ \frac{1}{r_A}-\frac{1}{r_B}=C,\qquad\text{equivalently}\qquad r_B-r_A=C\,r_Ar_B. \]

2. Checking whether this is a circle (it is not)

The source claimed these are “Apollonius circles” — the locus of a constant ratio \(r_B/r_A=\lambda\) — but the actual condition here is a constant difference of reciprocals, a genuinely different relation. To settle this, both radicals were cleared symbolically with Sympy: squaring \(r_B-r_A=Cr_Ar_B\) after isolating leaves \(r_A^2+r_B^2-C^2r_A^2r_B^2=2r_Ar_B\) (still containing one radical product), and squaring once more gives the fully polynomial equation \(\left(r_A^2+r_B^2-C^2r_A^2r_B^2\right)^2=4r_A^2r_B^2\) with \(r_A^2=x^2+y^2\), \(r_B^2=(x-a)^2+y^2\) substituted in. Expanding this in Sympy and computing its total degree in \(x,y\) gives degree \(8\), and \(\texttt{sympy.factor}\) shows it does not reduce to a product of low-degree (circle-type, degree-2) factors for generic \(a,C\). A relation whose fully-cleared algebraic form is an irreducible degree-8 curve cannot be a circle (degree 2); the “Apollonius circle” claim, and the boxed formula built on it (which used \(\lambda=e^{Ca}\), a quantity that belongs to the different, ratio-based problem), do not apply here and have been dropped.

3. The one genuinely simple special case

Setting \(C=0\) (i.e. \(V_0=0\)) reduces the relation to \(r_A=r_B\), i.e. equal distance from both charges — Sympy's \(\texttt{solve}\) on \(r_A=r_B\) confirms this is exactly the vertical line \(x=a/2\), the perpendicular bisector of \(AB\). This is the only equipotential line that is a straight curve; every other value of \(C\) gives a genuine (non-circular) quartic-type curve in \(r_A^2,r_B^2\), asymmetric about \(x=a/2\) since the two charges have opposite sign.

4. Qualitative shape

For \(C>0\) (potential of the same sign as being closer to \(+q\)) the curve \(1/r_A-1/r_B=C\) lies in the half-plane nearer \(A\), wrapping more tightly around \(A\) as \(C\) increases (approaching a small loop around \(+q\) as \(C\to\infty\)); by the antisymmetry of the defining relation under \((x,y)\mapsto(a-x,y)\), \(C\mapsto -C\), the \(C<0\) curves are the mirror images wrapping around \(-q\) near \(B\). Together with the bisector \(x=a/2\) at \(C=0\), these form the complete, correctly-derived family of equipotential lines — an implicit (non-circular) curve family, not the circle family the source asserted.