Physics · Ch 2 — Electrostatic Potential and Capacitance

Potential Due to an Electric Dipole

2.4

Potential Due to an Electric Dipole

Concept: Potential of a Dipole

An electric dipole consists of two equal and opposite charges $+q$ and $-q$ separated by a small distance $2a$ . Its total charge is zero, but it has a dipole moment vector $\mathbf{p}$ with magnitude $p = q \times 2a$ , pointing from $-q$ to $+q$ . Unlike a single point charge, the potential due to a dipole depends not only on the distance $r$ from its centre but also on the angle $\theta$ between $\mathbf{r}$ and $\mathbf{p}$ .

Derivation of Potential

We place the dipole at the origin, with its centre at $O$ . For a point $P$ with position vector $\mathbf{r}$ , the potential $V$ is the sum of potentials from each charge (superposition principle):

V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{r_1} - \frac{q}{r_2} \right)

where $r_1$ and $r_2$ are distances from $P$ to $+q$ and $-q$ , respectively.

From geometry (using the law of cosines):

r_1^2 = r^2 + a^2 - 2ar\cos\theta

r_2^2 = r^2 + a^2 + 2ar\cos\theta

For points far away ( $r \gg a$ ), we keep only first-order terms in $a/r$ . Using the binomial expansion:

\frac{1}{r_1} \approx \frac{1}{r} \left(1 + \frac{a}{r}\cos\theta\right)

\frac{1}{r_2} \approx \frac{1}{r} \left(1 - \frac{a}{r}\cos\theta\right)

Substituting into the expression for $V$ and using $p = 2qa$ :

V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2qa\cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p\cos\theta}{r^2}

Since $p\cos\theta = \mathbf{p} \cdot \hat{\mathbf{r}}$ (where $\hat{\mathbf{r}}$ is the unit vector along $\mathbf{r}$ ), the final result is:

\boxed{V = \frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^2}}

This formula is exact for a point dipole at the origin and approximately true for $r \gg a$ .

Special Cases

On the dipole axis ( $\theta = 0$ or $\theta = \pi$ ): $V = \pm \frac{1}{4\pi\varepsilon_0} \frac{p}{r^2}$ (positive for $\theta = 0$ , negative for $\theta = \pi$ ) …

Figure 2.5Quantities involved in the calculation of potential due to a dipole.

Fig. 2.5 — Quantities involved in the calculation of potential due to a dipole.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure shows a vertical electric dipole placed at the origin O. The positive charge $+q$ is at the top, the negative charge $-q$ at the bottom, and the separation between them is $2a$ . A vertical double-headed arrow on the left marks this distance, with the upper half labelled $a$ (from O to $+q$ ) and the lower half also labelled $a$ (from O to $-q$ ). At the centre O, a short bold arrow labelled p points upward — this is the dipole moment vector, defined as $\mathbf{p} = q \times 2a$ in the direction from $-q$ to $+q$ .

A point P is located at the upper-right, and three lines connect it to the dipole:

A dashed line from $+q$ to P, labelled $r_1$ (distance from the positive charge).
A dashed line from $-q$ to P, labelled $r_2$ (distance from the negative charge).
A bold line from O to P, labelled r (the position vector of P from the centre).

At O, the angle between the upward axis (direction of $\mathbf{p}$ ) and the line OP is marked $\theta$ . This angle is crucial because the potential depends on it.

Physical idea: The total potential at P is the sum of the potentials due to each charge (superposition principle). For a point far away ( $r \gg a$ ), the potential simplifies to a form that depends on both $r$ and $\theta$ , unlike the $1/r$ potential of a single charge.

Key formula derived from this figure:

V = \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^2}

where:

$p = 2qa$ is the magnitude of the dipole moment,
$\theta$ is the angle between $\mathbf{p}$ and $\mathbf{r}$ ,
$\hat{\mathbf{r}}$ is the unit vector along $\mathbf{r}$ , …

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