Two point charges of +Q each have been placed at the positions ( –a /2, 0, 0) and (a /2, 0, 0). The locus of the points where –Q charge can be placed such the that total electrostatic potential energy of the system can become equal to zero, is represented by which of the following equations?

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z² + (y–a)² = 27a² /4

z² + y² = 15a² /4

z² + (y–a)² = 2a

None

answer is C.

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Detailed Solution

PE will be zero when -Q lies in yz-plane.

Let P(0, y, z)

$\begin{aligned} U = 2 \times k \frac{Q (- Q)}{\sqrt{\frac{a^{2}}{4} + y^{2} + z^{2}}} + k \frac{(Q) Q}{a} = 0 \\ \Rightarrow 4 a^{2} = \frac{a^{2}}{4} + y^{2} + z^{2} \Rightarrow y^{2} + z^{2} = \frac{15 a^{2}}{4} \end{aligned}$

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