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Two charges q_{1} and q_{2} are placed at (0, 0, d) and (0, 0, -d) respectively. Find the locus of points where the potential is zero.

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We know that the potential at point P is V=\sum Vi

Where Vi=\frac{qi}{4\pi \varepsilon _{0}}, ri is the magnitude of the position vector P

V=\frac{1}{4 \pi \varepsilon _{0}}\sum \frac{qi}{rpi}

When the (x,y,z) plane is considered, the two charges lie on the z-axis and are separated by 2d. The potential is given as

\frac{q_{1}}{\sqrt{x^{2}+y^{2}+\left ( z-d \right )^{2}}}+\frac{q_{2}}{\sqrt{x^{2}+y^{2}+\left ( z+d \right )^{2}}}=0

Squaring the equation, we get

x^{2}+y^{2}+z^{2}+\left [ \left ( \frac{q_{1}^2+q_2^2}{q_{1}^2-q_2^2} \right )^2 \right ]\left ( 2zd \right )+d^2=0

The Centre of the sphere is

\left ( 0,0,-d\left [ \frac{q_{1}^{2}+q_{2}^{2}}{q_{1}^{2}-q_{2}^{2}} \right ] \right )

And radius is



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