Electric potential — point charge, dipole, conducting sphere formulas

Question

Write the electric potential formula for (a) a point charge, (b) an electric dipole at axial and equatorial points, and (c) a conducting sphere. At what point is the potential of a dipole zero?

(CBSE 12, JEE Main & NEET — very high frequency)

Solution — Step by Step

V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = \frac{kq}{r}

Potential is a scalar — no direction, just sign. Positive charge gives positive $V$ , negative charge gives negative $V$ . It decreases as $1/r$ (not $1/r^2$ like the field).

A dipole has charges $+q$ and $-q$ separated by distance $2a$ , with dipole moment $p = q \times 2a$ .

Axial point (on the line joining the charges, at distance $r \gg a$ ):

V_{\text{axial}} = \frac{kp}{r^2}

Equatorial point (on the perpendicular bisector):

V_{\text{equatorial}} = 0

General point (at angle $\theta$ from the axis):

V = \frac{kp\cos\theta}{r^2}

The potential is zero on the entire equatorial plane ( $\theta = 90°$ ) because the contributions from $+q$ and $-q$ cancel exactly.

Outside ( $r > R$ ): $V = \dfrac{kQ}{r}$ — behaves like a point charge

On the surface ( $r = R$ ): $V = \dfrac{kQ}{R}$

Inside ( $r < R$ ): $V = \dfrac{kQ}{R}$ — constant, same as surface

The interior is an equipotential region because $E = 0$ inside a conductor.

Why This Works

Potential measures the work done per unit charge to bring a test charge from infinity to that point. For a dipole, the equatorial point is equidistant from both charges, so the positive and negative contributions cancel — zero net work.

graph TD
    A["Electric Potential Problem"] --> B{"Charge distribution?"}
    B -->|"Point charge"| C["V = kq/r<br/>Falls as 1/r"]
    B -->|"Dipole"| D{"Position?"}
    B -->|"Conducting sphere"| E{"Location?"}
    D -->|"Axial"| F["V = kp/r²"]
    D -->|"Equatorial"| G["V = 0"]
    D -->|"General angle θ"| H["V = kp cosθ/r²"]
    E -->|"Outside (r > R)"| I["V = kQ/r"]
    E -->|"Surface or inside"| J["V = kQ/R (constant)"]

Alternative Method — Superposition for Dipole

Instead of memorising the dipole formula, compute potential at any point by superposing two point charges:

V = \frac{kq}{r_+} + \frac{k(-q)}{r_-} = kq\left(\frac{1}{r_+} - \frac{1}{r_-}\right)

For the equatorial point, $r_+ = r_- = \sqrt{r^2 + a^2}$ , so $V = 0$ . This approach works even when $r$ is not much larger than $a$ .

Remember the power of $r$ in the denominator: field falls as $1/r^2$ for a point charge and $1/r^3$ for a dipole. Potential falls as $1/r$ for a point charge and $1/r^2$ for a dipole. Potential always falls one power slower than field because $V = -\int E\,dr$ .

Common Mistake

Students confuse potential (scalar, $V$ ) with electric field (vector, $E$ ). At the equatorial point of a dipole, the potential is zero but the field is NOT zero — it points antiparallel to the dipole moment. Zero potential does not mean zero field. Similarly, inside a conducting sphere, $E = 0$ but $V \neq 0$ (it equals the surface potential). JEE Main 2023 had a question specifically testing this distinction.

Question

Solution — Step by Step

Why This Works

Alternative Method — Superposition for Dipole

Common Mistake

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