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Two-point charges of magnitude \text {+q and -q} are placed at \left ( \frac{-d}{2},0,0 \right ) and \left ( \frac{d}{2},0,0 \right ) respectively. Find the equation of the equipotential surface where the potential is zero.

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Here, the potential due to charges at the point P is

V_{p}=\frac{1}{4\pi\varepsilon _{0}}\frac{q}{r_{1}}+\frac{1}{4\pi\varepsilon _{0}}\frac{(-q)}{r_{2}}

The net electric potential at this point is zero,

Therefore, r_{1}=r_{2}

We know that

r_{1}=\sqrt{\left ( \frac{x+d}{2} \right )^{2}+h^{2}}

r_{2}=\sqrt{\left ( \frac{x-d}{2} \right )^{2}+h^{2}}

Solving the two equations, we the required equation in-plane x = 0 which is a y-z plane.

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