Potential due to an Electric Dipole 🚀

1 · Meet the Dipole ✨

An electric dipole has two equal-magnitude charges, \(+q\) and \(-q\), sitting \(2a\) apart. Its dipole moment points from \(-q\) to \(+q\) and has size \(p = q \times 2a\). :contentReference[oaicite:11]{index=11}

2 · Cooking Up the Potential 🔍

  1. Start with superposition. The total potential at a point \(P\) comes from adding the individual contributions: \[ V = \frac{1}{4\pi\varepsilon_0}\!\left(\frac{q}{r_1}-\frac{q}{r_2}\right). \] Here \(r_1\) and \(r_2\) are distances from \(P\) to \(+q\) and \(-q\). :contentReference[oaicite:12]{index=12}
  2. Relate the distances. Using the geometry in Fig. 2.5, \[ \begin{aligned} r_1^{2} &= r^{2}+a^{2}-2ar\cos\theta,\\ r_2^{2} &= r^{2}+a^{2}+2ar\cos\theta. \end{aligned} \] :contentReference[oaicite:13]{index=13}
  3. Simplify for far-away points. When \(r \gg a\), expand to first order in \(a/r\): \[ \frac{1}{r_1}\approx\frac{1}{r}\!\left(1-\frac{a}{r}\cos\theta\right),\quad \frac{1}{r_2}\approx\frac{1}{r}\!\left(1+\frac{a}{r}\cos\theta\right). :contentReference[oaicite:14]{index=14} \]
  4. Arrive at the compact form. Insert those reciprocals and note \(p=2qa\) to get \[ V=\frac{1}{4\pi\varepsilon_0}\,\frac{\mathbf p\!\cdot\!\hat{\mathbf r}}{r^{2}} \quad (r\gg a). \] :contentReference[oaicite:15]{index=15}

3 · Special Locations 🌟

On the dipole axis (\(\theta=0\) or \(\pi\)): \[ V = \pm\frac{1}{4\pi\varepsilon_0}\,\frac{p}{r^{2}} \] (positive in front of \(+q\), negative behind it). :contentReference[oaicite:16]{index=16}
On the equatorial plane (\(\theta=\tfrac{\pi}{2}\)): \(V = 0\). :contentReference[oaicite:17]{index=17}

4 · Patterns to Remember 📈

  • The potential depends on both the distance \(r\) and the angle \(\theta\) between \(\mathbf r\) and \(\mathbf p\). :contentReference[oaicite:18]{index=18}
  • It drops off as \(1/r^{2}\) (faster than the \(1/r\) for a single charge). :contentReference[oaicite:19]{index=19}
  • Rotate \(\mathbf r\) about \(\mathbf p\); points on that cone share the same potential—axial symmetry rules! :contentReference[oaicite:20]{index=20}

NEET High-Yield Highlights 🎯

  • Golden formula: \(V=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{\mathbf p\!\cdot\!\hat{\mathbf r}}{r^{2}}\).
  • Axis potential: \(+\dfrac{p}{4\pi\varepsilon_0 r^{2}}\) in front, \(-\dfrac{p}{4\pi\varepsilon_0 r^{2}}\) behind.
  • Equatorial plane gives zero potential—easy mark! 😉
  • Potential ∝ \(1/r^{2}\) vs. single charge’s \(1/r\): a frequent conceptual check.

Keep practicing—your understanding will dipole-vault to new heights! 🚀