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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

$r=\frac{2q_{1}q_{2}d}{q_{1}^{2}-q_{2}^{2}}$

Posted by

infoexpert23

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