Problem Statement
Two point charges $+q$ and $-q$ are separated by distance $2a$. Describe the equipotential surfaces and find the potential at the midpoint.
Given Information
- Charge $+q$ at position $(-a, 0)$
- Charge $-q$ at position $(+a, 0)$
- Separation $d = 2a$
Physical Concepts & Formulas
An equipotential surface is a locus of all points at the same electric potential. No work is done moving a charge along such a surface since $W = q\Delta V = 0$. For a single point charge, equipotentials are concentric spheres. For a dipole ($+q$ and $-q$), the superposition of two sets of spheres creates distorted surfaces — near each charge they resemble spheres, but far away they become more complex. The midpoint between the charges lies on the $V=0$ plane (the perpendicular bisector plane), because the contributions from $+q$ and $-q$ are equal in magnitude but opposite in sign.
- $V = \sum_i \dfrac{kq_i}{r_i}$ — scalar superposition of potentials
- $V_{+q} = \dfrac{kq}{r_1}$ where $r_1$ is distance from $+q$
- $V_{-q} = \dfrac{-kq}{r_2}$ where $r_2$ is distance from $-q$
- $V_{\text{midpoint}} = \dfrac{kq}{a} + \dfrac{k(-q)}{a} = 0$
Step-by-Step Solution
Step 1 — Place charges: Set $+q$ at $A(-a,0)$ and $-q$ at $B(+a,0)$.
Step 2 — Potential at midpoint O $(0,0)$:
$$V_O = \frac{kq}{a} + \frac{k(-q)}{a} = \frac{kq – kq}{a} = 0$$
Step 3 — Equipotential plane: The entire perpendicular bisector plane (the $yz$-plane here) is at $V=0$ because any point on it is equidistant from $+q$ and $-q$. This is called the zero-potential plane.
Step 4 — Shape of equipotentials: For $V = C > 0$, surfaces cluster around $+q$ and shrink toward it. For $V = C < 0$, they cluster around $-q$. Field lines run perpendicular to every equipotential surface.
Worked Calculation
Substituting all values with units:
At midpoint ($r_1 = r_2 = a$):
$$V = \frac{k(+q)}{a} + \frac{k(-q)}{a} = \frac{kq – kq}{a} = 0$$
At a general point $(x,y)$: $r_1 = \sqrt{(x+a)^2+y^2}$, $r_2 = \sqrt{(x-a)^2+y^2}$
$$V(x,y) = kq\left(\frac{1}{r_1} – \frac{1}{r_2}\right)$$
Answer
$$\boxed{V_{\text{midpoint}} = 0}$$
Physical Interpretation
The zero result at the midpoint is a direct consequence of the antisymmetry of the charge configuration — the $+q$ and $-q$ contributions exactly cancel. This point is NOT a field-free point; the electric field there is $E = 2kq/a^2$ directed from $+q$ toward $-q$. Equipotential surfaces and field lines always cross at right angles.
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