Q.Two charges $q_1$ and $q_2$ are placed at $(0, 0, d)$ and $(0, 0, -d)$ respectively. Find locus of points where the potential is zero.

Write $V=\frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)=0$ with $r_1^2=x^2+y^2+(z-d)^2$, $r_2^2=x^2+y^2+(z+d)^2$. Zero potential needs $q_1/r_1=-q_2/r_2$, which requires opposite-sign charges. Squaring gives $q_1^2 r_2^2=q_2^2 r_1^2$. If $|q_1|=|q_2|$ (i.e. $q_1=-q_2$) this reduces to the plane $z=0$; if $|q_1|\ne|q_2|$ it is a sphere. Like-sign charges have no zero-potential point. …

Concept understanding

Potential is a scalar, so the two contributions add algebraically:

with, for a general point $(x,y,z)$,

Setting $V=0$

Since $r_1,r_2>0$, this can hold only if $q_1$ and $q_2$ have opposite signs. Squaring (which may add extraneous roots, checked below):

Substituting and expanding gives the key relation, call it (*):

Cases

1. Equal and opposite, $q_1=-q_2$ (so $q_1^2=q_2^2$). The first bracket vanishes and (*) becomes $4zd\,q_1^2=0\Rightarrow z=0$. The locus is the $xy$-plane $z=0$, the perpendicular bisector where every point is equidistant from the two charges, so their potentials cancel.

2. Opposite signs, unequal magnitudes ($|q_1|\ne|q_2|$). Divide (*) by $(q_1^2-q_2^2)$ and complete the square in $z$:

using $(a+b)^2-(a-b)^2=4ab$. This is a sphere with centre $\left(0,0,\,z_0\right)$ and radius $R$: …

Method: Finding the Locus Where the Potential of a Point-Charge System Is Zero

Use this for problems that ask "where in space is $V=0$ (or some constant)" for a system of two or more point charges — the goal is a geometric surface, not a single point.

Steps

Step 1: Write the total potential as an algebraic sum

Because potential is a scalar, the potential from multiple charges simply adds (with sign), unlike field vectors that need components:

Express each $r_i$ in coordinates relative to that charge's actual position (e.g. $r_1 = \sqrt{x^2+y^2+(z-d)^2}$ for a charge at $(0,0,d)$).

Step 2: Set the sum equal to the target value and isolate one term

For a zero-potential locus with two charges, rearrange to $\dfrac{q_1}{r_1} = -\dfrac{q_2}{r_2}$. Since $r_1, r_2 > 0$ always, immediately check what this requires of the signs of the charges — this often rules out or restricts entire cases before any algebra (e.g. same-sign charges can never satisfy this).

Step 3: Square to remove the square roots, but track extraneous roots …