Prerequisites
Before studying this topic, make sure you understand:
- Electric Charges — Coulomb’s law and superposition
- Electric Field — Field due to point charges
- Basic vector addition and trigonometry
The Hook: Molecules Are Tiny Magnets!
Ever wonder why water dissolves salt but oil doesn’t? Or why a plastic comb can pick up tiny paper pieces after you comb your hair? Or in Oppenheimer (2023), how atoms interact to form molecules?
The secret is the electric dipole — most molecules are tiny electric dipoles! Water molecules have a positive end and a negative end, making them perfect at grabbing ions and charged particles. This simple concept explains chemistry, biology, and material science!
What is an Electric Dipole?
Key Features:
- Equal magnitude charges (usually denoted $q$)
- Opposite signs ($+q$ and $-q$)
- Small separation distance $2a$ (compared to observation distance)
- Net charge = 0 (but creates electric field!)
Examples in nature:
- Water molecule (H₂O) — oxygen side is negative, hydrogen side is positive
- HCl molecule — chlorine end is negative, hydrogen end is positive
- CO₂ — linear, so dipole moments cancel (not a dipole!)
- Atoms in external field — positive nucleus shifts one way, electron cloud the other
In movies: Think of Thor’s hammer Mjolnir — one end attracts (negative), one end repels (positive)!
Dipole Moment
The dipole moment is a measure of the “strength” of a dipole.
$$\boxed{\vec{p} = q \cdot 2a \cdot \hat{p}}$$Magnitude: $p = q \times 2a$
Direction: From negative charge to positive charge (conventional)
Unit: Coulomb-meter (C·m) or Debye (D) where 1 D = $3.33 \times 10^{-30}$ C·m
“Dipole points from Negative to Positive”
Think of it as the “shift” direction when you pull the charges apart — you pull the positive charge in the direction of $\vec{p}$.
Visual: Draw an arrow from $-q$ to $+q$ — that’s $\vec{p}$!
Why is dipole moment important?
- Larger $p$ → stronger interaction with fields
- Many molecular properties depend on dipole moment
- Dipole moment determines how molecules align in electric fields
Electric Field Due to a Dipole
The field pattern of a dipole is characteristic — curved lines from $+q$ to $-q$.
We’ll calculate field at two special positions:
- Axial line (along the dipole axis)
- Equatorial line (perpendicular bisector)
Conventions
- Center of dipole at origin
- Dipole aligned along an axis (say, $+q$ at $+a$, $-q$ at $-a$)
- Distance from center: $r$
- Assume $r >> a$ (observation point is far from dipole)
Field on the Axial Line
Position: Point P on the axis of dipole, at distance $r$ from center
Setup:
- Distance from $+q$: $(r - a)$
- Distance from $-q$: $(r + a)$
Field due to $+q$:
$$E_+ = k\frac{q}{(r-a)^2}$$(pointing away from $+q$, i.e., along positive axis)
Field due to $-q$:
$$E_- = k\frac{q}{(r+a)^2}$$(pointing toward $-q$, i.e., also along positive axis)
Net field:
$$E_{axial} = E_+ - E_- = kq\left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right]$$ $$E_{axial} = kq\left[\frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2}\right] = kq\left[\frac{4ar}{(r^2-a^2)^2}\right]$$For $r >> a$, we have $r^2 - a^2 \approx r^2$:
$$\boxed{E_{axial} = \frac{2kp}{r^3}}$$where $p = q \times 2a$ (dipole moment)
Direction: Along the dipole axis (from $-q$ to $+q$)
Alternative form:
$$\boxed{E_{axial} = \frac{2p}{4\pi\varepsilon_0 r^3}}$$Field falls off as $\frac{1}{r^3}$ (not $\frac{1}{r^2}$ like a point charge!)
This is because the dipole has net charge = 0. The two charges’ fields partially cancel, making the net field weaker.
Field on the Equatorial Line
Position: Point P on the perpendicular bisector, at distance $r$ from center
Setup:
- Distance from each charge: $\sqrt{r^2 + a^2}$
- Fields from $+q$ and $-q$ have equal magnitude but different directions
Field magnitude from each charge:
$$E_+ = E_- = k\frac{q}{r^2 + a^2}$$Direction analysis:
- Field from $+q$ points away from $+q$
- Field from $-q$ points toward $-q$
- By symmetry, components perpendicular to equatorial line cancel
- Components along dipole axis (toward $-q$) add
Component along axis:
$$E_{component} = E \cos\theta = E \frac{a}{\sqrt{r^2 + a^2}}$$Net field:
$$E_{equatorial} = 2 \times k\frac{q}{r^2 + a^2} \times \frac{a}{\sqrt{r^2 + a^2}} = \frac{2kqa}{(r^2 + a^2)^{3/2}}$$For $r >> a$:
$$\boxed{E_{equatorial} = \frac{kp}{r^3}}$$Direction: Parallel to dipole axis, from $+q$ to $-q$ (opposite to $\vec{p}$)
Alternative form:
$$\boxed{E_{equatorial} = \frac{p}{4\pi\varepsilon_0 r^3}}$$Comparing Axial and Equatorial Fields
| Property | Axial Field | Equatorial Field |
|---|---|---|
| Magnitude | $\frac{2kp}{r^3}$ | $\frac{kp}{r^3}$ |
| Direction | Along $\vec{p}$ | Opposite to $\vec{p}$ |
| Ratio | $E_{axial} : E_{equatorial}$ | $2 : 1$ |
| Distance dependence | $\frac{1}{r^3}$ | $\frac{1}{r^3}$ |
“Axial is TWICE Equatorial”
$E_{axial} = 2 \times E_{equatorial}$ (at same distance $r$)
“Axial ALigns, Equatorial Opposes”
- Axial field: same direction as $\vec{p}$
- Equatorial field: opposite to $\vec{p}$
Electric Field at General Point
For a point at distance $r$ making angle $\theta$ with the dipole axis:
$$\boxed{E = \frac{kp}{r^3}\sqrt{1 + 3\cos^2\theta}}$$Direction: Makes angle $\alpha$ with the axis where:
$$\tan\alpha = \frac{\tan\theta}{2}$$Special cases:
- $\theta = 0°$ (axial): $E = \frac{2kp}{r^3}$
- $\theta = 90°$ (equatorial): $E = \frac{kp}{r^3}$
Dipole in External Uniform Electric Field
Consider a dipole placed in a uniform external field $\vec{E}$.
Forces on the Dipole
Force on $+q$: $\vec{F}_+ = q\vec{E}$ (along $\vec{E}$)
Force on $-q$: $\vec{F}_- = -q\vec{E}$ (opposite to $\vec{E}$)
Net force:
$$\boxed{F_{net} = \vec{F}_+ + \vec{F}_- = 0}$$Key Point: Net force is zero (in uniform field) — dipole doesn’t accelerate!
But wait… if net force is zero, does nothing happen?
Torque on Dipole in Electric Field
Torque calculation:
The two forces form a couple (equal, opposite, parallel forces).
Torque magnitude: $\tau = \text{Force} \times \text{perpendicular distance}$
Perpendicular distance between forces = $2a \sin\theta$
$$\tau = qE \times 2a\sin\theta = (q \times 2a) E \sin\theta$$ $$\boxed{\tau = pE\sin\theta}$$Vector form:
$$\boxed{\vec{\tau} = \vec{p} \times \vec{E}}$$Direction: Tends to align $\vec{p}$ with $\vec{E}$ (clockwise or counterclockwise)
This is exactly like a compass needle (magnetic dipole) aligning with Earth’s magnetic field!
In Interstellar, when they use compasses on the ice planet — same principle, different type of dipole.
Stable equilibrium: $\vec{p}$ parallel to $\vec{E}$ ($\theta = 0°$, $\tau = 0$)
Unstable equilibrium: $\vec{p}$ antiparallel to $\vec{E}$ ($\theta = 180°$, $\tau = 0$)
Potential Energy of Dipole in Electric Field
Work done to rotate dipole from angle $\theta_1$ to $\theta_2$:
$$W = -\int_{\theta_1}^{\theta_2} \tau \, d\theta = -\int_{\theta_1}^{\theta_2} pE\sin\theta \, d\theta$$ $$W = pE[\cos\theta_2 - \cos\theta_1]$$Taking reference ($U = 0$) at $\theta = 90°$:
$$\boxed{U = -pE\cos\theta}$$Vector form:
$$\boxed{U = -\vec{p} \cdot \vec{E}}$$Energy States
| Orientation | Angle $\theta$ | Energy $U$ | Stability |
|---|---|---|---|
| Parallel to $\vec{E}$ | $0°$ | $-pE$ | Stable (minimum) |
| Perpendicular to $\vec{E}$ | $90°$ | $0$ | Reference |
| Antiparallel to $\vec{E}$ | $180°$ | $+pE$ | Unstable (maximum) |
- Minimum energy ($-pE$): Dipole aligned with field (stable)
- Maximum energy ($+pE$): Dipole opposite to field (unstable)
- Work to rotate 180°: $W = 2pE$
This is why dipoles naturally align with electric fields — they seek minimum energy!
Dipole in Non-Uniform Electric Field
If field is not uniform, the forces on $+q$ and $-q$ are different.
Net force:
$$\vec{F}_{net} \neq 0$$Result: Dipole experiences both:
- Translational motion (due to net force)
- Rotational motion (due to torque)
Direction of motion: Toward region of stronger field
This is how charged particles are trapped in electromagnetic traps!
Electric Potential Due to Dipole
At a point distance $r$ from center, making angle $\theta$ with axis:
$$\boxed{V = \frac{kp\cos\theta}{r^2}}$$Vector form:
$$V = \frac{k\vec{p} \cdot \hat{r}}{r^2}$$Special cases:
- On axis ($\theta = 0°$): $V = \frac{kp}{r^2}$
- On equatorial line ($\theta = 90°$): $V = 0$
Wrong: “Equatorial field is zero”
Correct:
- Equatorial potential is zero
- Equatorial field is NOT zero: $E = \frac{kp}{r^3}$
Don’t confuse $V$ and $E$!
Memory Tricks & Patterns
Mnemonic for Dipole Formulas
“Two Kinds Please” → $E_{axial} = \frac{2kp}{r^3}$
“King’s Palace” → $E_{equatorial} = \frac{kp}{r^3}$
“Positive Equals Torque” → $\tau = pE\sin\theta$
“Potential Comes Over Radius-squared” → $V = \frac{kp\cos\theta}{r^2}$
Pattern Recognition
| Quantity | Dependence on $r$ |
|---|---|
| Point charge field | $E \propto \frac{1}{r^2}$ |
| Dipole field | $E \propto \frac{1}{r^3}$ |
| Point charge potential | $V \propto \frac{1}{r}$ |
| Dipole potential | $V \propto \frac{1}{r^2}$ |
Rule: Dipole quantities fall off one power of r faster than point charge!
When to Use Dipole Approximation
Use dipole formulas when:
- Two equal and opposite charges
- Observation point is far: $r >> 2a$ (distance between charges)
- Problem mentions “electric dipole”
Don’t use dipole approximation when:
- Charges are not equal and opposite
- Observation point is close to the charges
- Need exact field (not approximate)
For exact calculations near dipole: Use superposition with individual charges!
Common Mistakes to Avoid
Wrong: $\vec{p}$ points from positive to negative
Correct: $\vec{p}$ points from negative to positive
Remember: It’s the “displacement” of positive charge!
Wrong: “Dipole in uniform field experiences net force”
Correct:
- Uniform field: Net force = 0, but torque exists
- Non-uniform field: Both force and torque exist
Torque rotates, force translates!
Wrong: “Equatorial field is zero, so potential is also zero”
Correct:
- Equatorial potential = 0 (correct!)
- Equatorial field ≠ 0 (it’s $\frac{kp}{r^3}$)
Zero potential doesn’t mean zero field!
Wrong: $U = pE\cos\theta$
Correct: $U = -pE\cos\theta$ (negative sign!)
Minimum energy is at $\theta = 0°$ (aligned), which gives $U = -pE$ (negative).
Practice Problems
Level 1: Foundation (NCERT)
Two charges $+10$ μC and $-10$ μC are separated by 4 cm. Find the dipole moment.
Solution:
$$p = q \times 2a = 10 \times 10^{-6} \times 4 \times 10^{-2}$$ $$\boxed{p = 4 \times 10^{-7} \text{ C·m}}$$Direction: From negative to positive charge
A dipole of moment $4 \times 10^{-9}$ C·m is placed in a uniform field of $5 \times 10^5$ N/C. Find the maximum torque.
Solution:
Maximum torque occurs at $\theta = 90°$ (perpendicular to field):
$$\tau_{max} = pE\sin 90° = pE$$ $$\tau_{max} = 4 \times 10^{-9} \times 5 \times 10^5$$ $$\boxed{\tau_{max} = 2 \times 10^{-3} \text{ N·m} = 2 \text{ mN·m}}$$Find the electric field on the axis of a dipole of moment $10^{-8}$ C·m at a distance of 10 cm from its center.
Solution:
$$E_{axial} = \frac{2kp}{r^3} = \frac{2 \times 9 \times 10^9 \times 10^{-8}}{(0.1)^3}$$ $$E_{axial} = \frac{18 \times 10}{10^{-3}} = 1.8 \times 10^5 \text{ N/C}$$ $$\boxed{E = 1.8 \times 10^5 \text{ N/C}}$$Level 2: JEE Main
A dipole is placed at the origin with its axis along the y-axis. If the electric field at point $(0, d, 0)$ is $E$, find the field at point $(d, 0, 0)$.
Solution:
Point $(0, d, 0)$ is on the axis (y-axis).
$$E_{axial} = \frac{2kp}{d^3} = E$$So, $kp = \frac{Ed^3}{2}$
Point $(d, 0, 0)$ is on the equatorial line (x-axis).
$$E_{equatorial} = \frac{kp}{d^3} = \frac{Ed^3/2}{d^3} = \frac{E}{2}$$ $$\boxed{E_{equatorial} = \frac{E}{2}}$$Key: Equatorial field is half of axial field!
Work done in rotating a dipole from $\theta = 0°$ to $\theta = 60°$ in a uniform field is $W$. Find the work to rotate from $60°$ to $120°$.
Solution:
Work done = Change in potential energy:
$$W = U_f - U_i = -pE\cos\theta_f - (-pE\cos\theta_i) = pE(\cos\theta_i - \cos\theta_f)$$First rotation ($0°$ to $60°$):
$$W_1 = pE(\cos 0° - \cos 60°) = pE(1 - 0.5) = 0.5pE = W$$So, $pE = 2W$
Second rotation ($60°$ to $120°$):
$$W_2 = pE(\cos 60° - \cos 120°) = pE(0.5 - (-0.5)) = pE$$ $$\boxed{W_2 = 2W}$$At what angle does the electric field of a dipole make an angle of $45°$ with the dipole axis?
Solution:
Field direction makes angle $\alpha$ with axis where:
$$\tan\alpha = \frac{\tan\theta}{2}$$Given $\alpha = 45°$:
$$\tan 45° = \frac{\tan\theta}{2}$$ $$1 = \frac{\tan\theta}{2}$$ $$\tan\theta = 2$$ $$\boxed{\theta = \tan^{-1}(2) \approx 63.43°}$$Level 3: JEE Advanced
Two dipoles of moments $p_1$ and $p_2$ are placed on the x-axis at $x = -a$ and $x = +a$ respectively, both pointing in the +y direction. Find the electric field at the origin.
Solution:
Both dipoles create field at origin (which is on their equatorial line).
For dipole at $x = -a$:
- Distance from origin: $a$
- Equatorial field: $E_1 = \frac{kp_1}{a^3}$ pointing in -y direction (opposite to dipole)
For dipole at $x = +a$:
- Distance from origin: $a$
- Equatorial field: $E_2 = \frac{kp_2}{a^3}$ pointing in -y direction
Both fields are in the same direction (-y)!
$$E_{net} = E_1 + E_2 = \frac{k(p_1 + p_2)}{a^3}$$ $$\boxed{E = \frac{k(p_1 + p_2)}{a^3} \text{ in -y direction}}$$A dipole of moment $p$ is placed at a distance $r$ from a point charge $Q$. The dipole is oriented radially (along the line joining them). Find the force on the dipole.
Solution:
Field due to $Q$ at distance $r$: $E = \frac{kQ}{r^2}$
This field is non-uniform (changes with position).
Force on dipole in non-uniform field:
$$F = p\frac{dE}{dr}$$ $$\frac{dE}{dr} = \frac{d}{dr}\left(\frac{kQ}{r^2}\right) = kQ \times (-2)r^{-3} = -\frac{2kQ}{r^3}$$ $$F = p \times \frac{2kQ}{r^3}$$ $$\boxed{F = \frac{2kpQ}{r^3}}$$Direction: Toward $Q$ if $Q > 0$, away if $Q < 0$
Note: Force exists because field is non-uniform!
Show that the electric field of a dipole at large distances is perpendicular to the line joining the observation point to the dipole at $\theta = \tan^{-1}(\sqrt{2})$.
Solution:
Field at angle $\theta$ makes angle $\alpha$ with axis where:
$$\tan\alpha = \frac{\tan\theta}{2}$$For field to be perpendicular to radial direction, $\alpha + \theta = 90°$:
$$\alpha = 90° - \theta$$ $$\tan(90° - \theta) = \frac{\tan\theta}{2}$$ $$\cot\theta = \frac{\tan\theta}{2}$$ $$\frac{1}{\tan\theta} = \frac{\tan\theta}{2}$$ $$\tan^2\theta = 2$$ $$\boxed{\theta = \tan^{-1}(\sqrt{2}) \approx 54.74°}$$At this “magic angle,” the field is tangent to a circle centered at the dipole!
Quick Revision Box
| Concept | Formula/Key Point |
|---|---|
| Dipole moment | $\vec{p} = q \times 2a$ (from - to +) |
| Axial field | $E = \frac{2kp}{r^3}$ (along $\vec{p}$) |
| Equatorial field | $E = \frac{kp}{r^3}$ (opposite to $\vec{p}$) |
| Field at angle $\theta$ | $E = \frac{kp}{r^3}\sqrt{1 + 3\cos^2\theta}$ |
| Torque in uniform field | $\tau = pE\sin\theta$ or $\vec{\tau} = \vec{p} \times \vec{E}$ |
| Potential energy | $U = -pE\cos\theta$ or $U = -\vec{p} \cdot \vec{E}$ |
| Potential at angle $\theta$ | $V = \frac{kp\cos\theta}{r^2}$ |
| Axial : Equatorial field | $2 : 1$ (at same distance) |
| Stable orientation | $\vec{p} \parallel \vec{E}$ ($U = -pE$) |
| Unstable orientation | $\vec{p}$ antiparallel to $\vec{E}$ ($U = +pE$) |
JEE Weightage: 1-2 questions in JEE Main (formula-based), 1 question in JEE Advanced (often combined with torque/energy)
Time-Saver: Remember the 2:1 ratio for axial:equatorial — saves calculation time!
Teacher’s Summary
- Dipole moment direction: From negative to positive charge (conventional)
- Field falls as 1/r³ — faster than point charge because net charge = 0
- Axial field is twice equatorial field — important ratio to remember
- Torque aligns dipole with field — even though net force is zero (uniform field)
- Minimum energy when aligned — dipoles naturally align with electric field
“The dipole is the simplest charge distribution with net charge zero. It’s the building block for understanding molecules and materials.”
Related Topics
Within Electrostatics
- Electric Charges — Understanding individual charges
- Electric Field — Field due to charges
- Gauss’s Law — Alternative approach to fields
- Electric Potential — Potential due to dipole
Connected Chapters
- Magnetism — Magnetic dipole (current loop) is analogous
- Electromagnetic Induction — Rotating dipoles in fields
- Atoms and Nuclei — Atomic dipoles in spectroscopy
Chemistry Connections
- Molecular polarity — HCl, H₂O are dipoles
- Intermolecular forces — Dipole-dipole interactions
- Solubility — “Like dissolves like” (polar in polar)
Math Connections
- Vectors — Dot product, cross product for $U$ and $\tau$
- Trigonometry — Angular dependence
- Calculus — Deriving field from potential