Two point charges of magnitude + q and - q are placed at `(-(d)/(2),0,0)` and `((d)/(2),0,0)` . respectively. Find the equation of the equipotential surface where the potential is zero.

Explanation

Let the required plane lies at a distance x from the origin as shown in figure.

Potential at point P
`(kp)/([(x+(d)/(2))^(2)+h^(2)]^(1//20))-(kq)/([(x-(d)/(2))^(2)+h^(2)]^(1//2))=0`
`:. (1)/([(x+(d)/(2))^(2)+h^(2)]^(1//2))=(1)/([(x-(d)/(2))^(2)+h^(2)]^(1//2))`
`:. (x+(d)/(2))^(2)+h^(2)=(x+(d)/(2))^(2)+h^(2)`
`:. x^(2)-xd+(d^(2))/(4)=x^(2)+xd+(d^(2))/(4)`
`:. 0 =2xd`
`:. x=0`
It is the required equation of the potential for equipotential surface at x = 0 means yz plane.