Two point charges of magnitude $+q$ 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.
Two point charges of magnitude $+q$ 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.
Let the required plane lies at a distance $x$ from the origin as shown in figure.
Potential at point $P,$
$\frac{k q}{\left[\left(x+\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}-\frac{k q}{\left[\left(x-\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}=0$
$\therefore \frac{1}{\left[\left(x+\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}=\frac{1}{\left[\left(x-\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}$
$\therefore\left(x-\frac{d}{2}\right)^{2}+h^{2}=\left(x+\frac{d}{2}\right)^{2}+h^{2}$
$\therefore x^{2}-x d+\frac{d^{2}}{4}=x^{2}+x d+\frac{d^{2}}{4}$
$\therefore 0=2 x d$
$\therefore x=0$
The equation of the required plane is $x=0$ means $y z$ plane.
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- [AIIMS 2015]