Question
Two electric charges, +q at the origin O (0, 0) and −2q at the point A (6, 0) are placed on x axis. The locus of the point P in x-y plane where the potential vanishes (V = 0) is
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Solution
[Key Concept: The zero-potential locus of two unlike unequal point charges is the circle on which the distance ratio to the charges is fixed (Apollonius circle).]
$$V_{(x,y)} = 0 = \frac{kq}{\sqrt{x^{2} + y^{2}}} - \frac{k2q}{\sqrt{(x - 6)^{2} + y^{2}}}$$
$$\begin{matrix} & \ \Rightarrow (x - 6)^{2} + y^{2} = 4\left( x^{2} + y^{2} \right) \\ & \ \Rightarrow x^{2} + y^{2} + 36 - 12x = 4x^{2} + 4y^{2} \\ & \ \Rightarrow 3x^{2} + 3y^{2} + 12x - 36 = 0 \\ & \ \Rightarrow x^{2} + y^{2} + 4x - 12 = 0 \\ & \ \Rightarrow (x + 2)^{2} + y^{2} = 16\ldots\ldots\text{~Equation of circle~}. \end{matrix}$$
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