Potential of a Dipole and System of Charges

For a dipole, the potential has a beautiful $\cos\theta/r^2$ structure. And because potential is scalar, the potential of a many-charge system is the easiest electrostatic quantity to compute.

Concept

For a dipole with moment $\vec p$ centred at the origin, the potential at a point $\vec r$ (with $r\gg a$) is $V(r,\theta) = \frac{1}{4\pi\varepsilon_0}\frac{p\cos\theta}{r^2} = \frac{\vec p\cdot \hat r}{4\pi\varepsilon_0 r^2}.$ Where $\theta$ is the angle between $\vec p$ and the position vector.

Special cases:

• Axial point ($\theta = 0$): $V_\text{axial} = +kp/r^2$.
• Equatorial point ($\theta = \pi/2$): $V_\text{eq} = 0$.
• Behind axial ($\theta = \pi$): $V = -kp/r^2$.

Note: dipole potential falls as $1/r^2$, faster than a point charge's $1/r$.

System of charges. For $N$ charges at positions $\vec r_i$, the potential at $\vec r$ is $V(\vec r) = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{N}\frac{q_i}{|\vec r - \vec r_i|}.$ Algebraic sum — no vectors.

Derivation

Place $+q$ at $+a\hat z$ and $-q$ at $-a\hat z$. At point $\vec r$ with polar angle $\theta$ from the dipole axis: $V = kq\left(\frac{1}{r_+} - \frac{1}{r_-}\right),$ where $r_\pm$ is the distance from the field point to $\pm q$. For $r\gg a$ one can expand $r_\pm \approx r \mp a\cos\theta.$ Then $\frac{1}{r_+} - \frac{1}{r_-} \approx \frac{1}{r-a\cos\theta} - \frac{1}{r+a\cos\theta} \approx \frac{2a\cos\theta}{r^2}.$ With $p = 2aq$: $V \approx \frac{kp\cos\theta}{r^2}.$

Worked Example

A dipole of moment $p = 2\times 10^{-9}\,C\cdot m$ is centred at origin along $\hat z$. Find $V$ at a point $20\,cm$ from the centre at $\theta = 60^\circ$ from the axis.

$V = \frac{kp\cos\theta}{r^2} = \frac{9\times 10^9 \times 2\times 10^{-9} \times 0.5}{(0.20)^2} = \frac{9}{0.04} = 225\,V.$

For a system: three charges $+2, -1, +3\,nC$ at distances $0.10, 0.15, 0.25\,m$ from a point: $V = k\left(\frac{2}{0.10} - \frac{1}{0.15} + \frac{3}{0.25}\right)\times 10^{-9} = 9\times 10^9\,(20 - 6.67 + 12)\times 10^{-9} = 228\,V.$

Common Confusions

• On the equatorial plane the potential is zero — but the field is not zero ($E = kp/r^3$).
• Far-field formula assumes $r\gg a$. For close-up problems, use the exact superposition formula.
• Dipole $V$ falls as $1/r^2$, not $1/r$. Memorize this.
• Don't forget the cosine. On the equatorial line, $\cos\theta = 0$ and $V = 0$.

Key Takeaways

• Dipole potential: $V = kp\cos\theta/r^2 = \vec p\cdot\hat r/(4\pi\varepsilon_0 r^2)$.
• Axial: $+kp/r^2$. Equatorial: $0$. Behind: $-kp/r^2$.
• For systems: $V = \sum_i kq_i/r_i$.
• Scalar addition is much simpler than vector addition.