Two electric charges `q and - 2 q` are placed at a distance 6 m apart on a horizontal plane. Find the locus of point on this plane where the potential has a value zero.
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Explanation

If point P is at a distance `r_(1)` from point charge +q and `r_(2)` from charge `-2 q( Fig . 3.8)`, then
`V_P = V_1 + V_2 = (1)/(4 pi epsilon _0)[(+q )/r_(1)+(-2 q)/r_(2)]`
According to the problem, `V_P = 0, i.e.,`
`q/r_(1) - (2 q)/r_(2) = 0`
or `r_2 = 2r_1` (i)
But if the charge q is assumed to be situated at the origin, then
`r_1^2 = x^2 + y^2 and r_2 ^2 = (6 -x)^2 + y^2`
So substituting these values of `r_(1)` and `r_(2)` in Eq. (i), we get
`(x + 2)^2 + y^2 = 16`
So the locus of the point P is a circle with radius 4 m and center `(- 2, 0)`.