Potential at a Point Due to an Electric Dipole

1. Statement of the Concept

The electric potential at any point due to an electric dipole is defined as the algebraic sum of the potentials produced by the two equal and opposite charges constituting the dipole.

The dipole consists of charges +q and –q separated by a small distance 2a.

The potential depends on:

• the dipole moment [\vec{p} = 2aq]
• the distance of the point from the dipole
• the angle the point makes with the dipole axis.

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2. Clear Explanation and Mathematical Derivation

Basic Geometry

Let the dipole have:

• Charges: [+q] and [−q]
• Separation: [2a]
• Dipole moment:
  [
  \vec{p} = 2aq , \hat{p}
  ]

Let point P be located at a distance [r] from the center of the dipole, making an angle [\theta] with the dipole axis.

 

Distances of P from charges are:

• From +q: [r_-]
• From –q: [r_+]

Potential Due to Dipole

Potential at P:

[V] [= \dfrac{1}{4\pi\varepsilon_0} \left( \dfrac{+q}{r_-} – \dfrac{q}{r_+} \right)]

For points far from the dipole ([r \gg a]), using binomial approximation:

[
r_- = r – a\cos\theta
]
[
r_+ = r + a\cos\theta
]

Then,

[V] [\approx \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q}{r – a\cos\theta}\dfrac{q}{r + a\cos\theta}\right)]

Simplifying using binomial expansion:

[V] [\approx \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{2qa\cos\theta}{r^2}\right)]

Since [p = 2aq]:

[V] [= \dfrac{1}{4\pi\varepsilon_0} \dfrac{p\cos\theta}{r^2}]

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3. Dimensions and Units

• Dimensional Formula:
  [V] = [M L^2 T^{-3} A^{-1}]
• SI Unit: Volt (V)

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4. Key Features

1. Potential depends on angle [\theta] between dipole moment and point P.
2. Potential falls off as [\dfrac{1}{r^2}], faster than point charge potential (which falls as [\dfrac{1}{r}]).
3. Potential is positive on the axial line in the direction of dipole moment.
4. Potential is zero on the equatorial line ([\theta = 90^\circ]).
5. For far-field points ([r \gg a]), potential depends only on dipole moment p, not on individual charges.
6. Dipoles are important in molecules, insulators, polarization, and dielectric theory.

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5. Important Formulas to Remember

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6. Conceptual Questions With Solutions

Q1. Why does potential due to a dipole fall as [\dfrac{1}{r^2}]?

**Solution:** Because the potentials of +q and –q almost cancel each other at large distances. Their first-order difference gives a [\dfrac{1}{r^2}] dependence.

Q2. Why is the potential zero at the equatorial line?

**Solution:** Because contributions from +q and –q are equal in magnitude but opposite in sign at [\theta = 90^\circ].

Q3. Is potential a vector for dipoles?

**Solution:** No, potential is scalar. Only depends on cosθ, not direction.

Q4. What determines sign of potential on the axial line?

**Solution:** The direction of dipole moment. If point lies in direction of p, potential is positive.

Q5. Can dipole still produce potential if charges are far apart?

**Solution:** Yes, the exact expression works, but the far-field formula works only when [r \gg a].

Q6. Why is the simplified formula not valid close to the dipole?

**Solution:** Because binomial approximation fails; exact distances r₋ and r₊ must be used.

Q7. Does potential become infinite at dipole center?

**Solution:** No. At the center, distances to charges are finite → finite potential.

Q8. Do dipoles create electric field even if potential is zero?

**Solution:** Yes. On equatorial line, potential is zero but field is non-zero.

Q9. Can potential for dipole be negative?

**Solution:** Yes. When [\cos\theta] is negative, V becomes negative.

Q10. Why is dipole moment important?

**Solution:** It characterizes the dipole’s strength; all far-field effects depend on p, not on q and a individually.

Q11. If charges are doubled but separation is halved, what happens to dipole moment?

**Solution:** [p = 2aq] → if q doubles and a halves, p stays same.

Q12. How does angle affect potential?

**Solution:** V ∝ cosθ. Maximum on axis, zero on equator, negative on opposite axis.

Q13. Can a single charge be treated as a dipole?

**Solution:** No. A dipole requires two charges of equal magnitude and opposite sign.

Q14. Why is potential larger on axial than equatorial line?

**Solution:** Because cosθ = 1 on axis, but cosθ = 0 on equator.

Q15. Is dipole potential continuous everywhere?

**Solution:** Yes, except at the exact positions of +q and –q.

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7. FAQ / Common Misconceptions

1. “Dipole potential decreases as 1/r like point charge.”

No. It decreases as [1/r^2].

2. “Dipole potential is always positive.”

It can be negative depending on angle.

3. “Potential and field both become zero on equatorial line.”

False. Only potential is zero; field is non-zero.

4. “Dipole moment depends only on separation, not charge.”

Incorrect. [p = 2aq], depends on both.

5. “Dipole potential is same everywhere on spherical shell around dipole.”

No. It depends on θ (cosine term).

6. “If dipole moment is zero, potentials still exist.”

No. Zero p means no net dipole → V becomes zero in far field.

7. “Dipole potential is constant on axial line.”

No, it drops as [1/r^2].

8. “Dipole must be physical charges separated by distance.”

Not always; molecules have *effective* dipole moments.

9. “Dipole potential is symmetric like point charge.”

Incorrect; it is direction-dependent (cosθ).

10. “Dipole potential equals zero at dipole center.”

False. Contributions do not cancel.

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8. Practice Questions (with Step-by-Step Solutions)

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Q1.

Dipole moment [p = 3 × 10^{-8} C·m].
Find potential at a point 20 cm away on axial line.

Solution:
Given: [r = 0.2 m], [\cos\theta = 1]

[V] [= \dfrac{1}{4\pi\varepsilon_0} \dfrac{p}{r^2}]

[V] [= 9\times10^9 \times \dfrac{3\times10^{-8}}{(0.2)^2}]

[V] [= 9\times10^9 \times \dfrac{3\times10^{-8}}{0.04}] [= 6.75 \times 10^{4} \text{V}]

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Q2.

Find potential on equatorial line at distance r from dipole of moment p.

Solution:
Equatorial → [\theta = 90^\circ], ⇒ [\cos\theta = 0]

Thus [V = 0].

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Q3.

Dipole consists of charges [+5 μC] and [–5 μC] separated by [6 cm].
Find potential at a far point where r = 60 cm and θ = 60°.

Solution:

Dipole moment:
[p = 2aq] [= 2(0.03)(5\times10^{-6})] [= 3\times10^{-7} \text{C·m}]

[V] [= \dfrac{1}{4\pi\varepsilon_0} \dfrac{p\cos\theta}{r^2}]

[V] [= 9\times10^9 \times \dfrac{3\times10^{-7}\times \dfrac{1}{2}}{(0.6)^2}]

[V] [= 9\times10^9 \times \dfrac{1.5\times10^{-7}}{0.36}]

[V = 3.75\times10^{3} \text{V}]

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Q4.

If dipole moment is doubled but distance from point is also doubled, what happens to potential?

Solution:
Original:
[
V \propto \dfrac{p}{r^2}
]

New:
[V’ \propto \dfrac{2p}{(2r)^2}] [= \dfrac{2p}{4r^2}] [= \dfrac{p}{2r^2}]

Thus:
[
V’ = \dfrac{1}{2}V
]

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Q5.

Charges [+q] and [−q] separated by 2a produce potentials [V_+] and [V_-] at a point.
Show that total potential is proportional to cosθ.

Solution:

[V] [= \dfrac{1}{4\pi\varepsilon_0}\left( \dfrac{q}{r_-} – \dfrac{q}{r_+} \right)]

Using approximations:

[r_- = r – a\cos\theta] [,\quad] [r_+ = r + a\cos\theta]

Simplifying:

[V] [= \dfrac{p\cos\theta}{4\pi\varepsilon_0 r^2}]

Thus V ∝ cosθ.