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Kumar Rohan

Physics and Mathematics

Dipole in a Uniform Electric Field

1. Statement of the Concept / Overview

A dipole (two equal and opposite charges [+q] and [-q] separated by distance [2a]) placed in a uniform electric field [\vec{E}] experiences a torque that tends to align its dipole moment [\vec{p}] with [\vec{E}]. In a uniform field there is no net force on the dipole (only torque), but there is potential energy associated with its orientation.

2. Clear Explanation and Mathematical Derivation

Consider a dipole with dipole moment:

[\vec{p} = q(2a),\hat{n}]

Place the dipole in a uniform electric field [\vec{E}]. Let the angle between [\vec{p}] and [\vec{E}] be [\theta] (measured from [\vec{p}] to [\vec{E}]).

(A) Forces on the Charges

Force on [+q]: [\vec{F}_+ = q\vec{E}].
Force on [-q]: [\vec{F}_- = -q\vec{E}].

These two forces are equal in magnitude and opposite in direction, so the net force is:

[\vec{F}{\text{net}}] [= \vec{F}+ + \vec{F}_-] [= q\vec{E} – q\vec{E}] [= \vec{0}]

Thus, no net translation in a uniform field.

Dipole in a Uniform Electric Field - Ucale — Image Credit: Ucale.org

(B) Torque on the Dipole

Take moments about the dipole centre. The two forces form a couple producing torque. Magnitude of torque:

[\tau = pE\sin\theta]

Vector form (tending to align [\vec{p}] with [\vec{E}]):

[\vec{\tau} = \vec{p} \times \vec{E}]

Direction given by right-hand rule.

Derivation (short)

Each force produces moment: [F = qE], lever arm = [a\sin\theta] for each charge → total torque:

[\tau] [= 2 \times (qE)(a\sin\theta)] [= 2aqE\sin\theta] [= pE\sin\theta]

(C) Potential Energy of a Dipole in Uniform Field

Work done to rotate dipole from reference angle [\theta_0] to [\theta]:

[W] [= \int_{\theta_0}^{\theta} \tau,d\theta] [= \int_{\theta_0}^{\theta} pE\sin\theta,d\theta]

Choosing reference [\theta_0 = 90^\circ] (often) or general expression:

Integrate:

[U(\theta)] [= -\int \tau,d\theta] [= -\int pE\sin\theta,d\theta] [= -pE\cos\theta + C]

Choosing zero of potential energy at [\theta = 90^\circ] gives [C = 0] and:

[U(\theta) = -pE\cos\theta]

Commonly used full expression (zero chosen at [\theta=90^\circ]):

Minimum (stable) energy at [\theta = 0^\circ]: [U_{\min} = -pE].
Maximum (unstable) energy at [\theta = 180^\circ]: [U_{\max} = +pE].

(D) Small Oscillations about Stable Equilibrium

If dipole is slightly displaced about [\theta = 0], torque ≈ [-pE\theta] (for small [\theta] where [\sin\theta \approx \theta]). Equation of motion:

[I \ddot{\theta} = -pE\theta]

where [I] is moment of inertia → simple harmonic motion with angular frequency:

[\omega = \sqrt{\dfrac{pE}{I}}]

3. Dimensions and Units

Quantity	Symbol	SI unit	Dimensions
Dipole moment	[p]	C·m	([I T L])
Electric field	[E]	N/C or V/m	([M L T^{-3} I^{-1}])
Torque	[\tau]	N·m	([M L^2 T^{-2}])
Potential energy	[U]	J	([M L^2 T^{-2}])

4. Key Features

Net force = 0 in a uniform field (no translation).
Torque = [pE\sin\theta] tends to align dipole with field.
Potential energy = [-pE\cos\theta]; lowest when [\vec{p}] // [\vec{E}] (aligned).
Dipole behaves like a tiny compass in an electric field (analogous to magnetic dipole in B-field).
In a non-uniform field, dipole experiences a net force in addition to torque.
For small angular displacements, dipole undergoes simple harmonic motion about stable equilibrium.

5. Important Formulas to Remember

Concept	Formula
Force on each charge	[F = qE]
Net force (uniform field)	[\vec{F}_{\text{net}} = \vec{0}]
Torque (magnitude)	[\tau = pE\sin\theta]
Torque (vector)	[\vec{\tau} = \vec{p} \times \vec{E}]
Potential energy	[U(\theta) = -pE\cos\theta]
Small oscillation frequency	[\omega = \sqrt{\dfrac{pE}{I}},]
Work to rotate from [\theta_1] to [\theta_2]	[W] [= -pE(\cos\theta_2 – \cos\theta_1)]

6. Conceptual Questions with Solutions

1. Why does a dipole experience no net force in a uniform electric field?

Because forces on +q and −q are equal in magnitude and opposite in direction, so they cancel: \[\vec{F}_+ = q\vec{E},\ \vec{F}_- = -q\vec{E}\], hence \[\vec{F}_{\text{net}} = \vec{0}\].

2. Why does a dipole experience a torque in a uniform field?

Although forces cancel, they act at different points and form a couple; this couple produces a torque \[\tau = pE\sin\theta\] that tends to rotate the dipole to align \[\vec{p}\] with \[\vec{E}\].

3. Why is potential energy lowest when dipole aligns with [\vec{E}]?

Because then \[\cos\theta = 1\] and \[U = -pE\] is minimal; the dipole has done work aligning with the field and system energy decreases.

4. If net force is zero, how can dipole move?

In a uniform field it cannot translate due to zero net force; it can only rotate due to torque. In a non-uniform field net force can be non-zero and cause translation.

5. Why is torque proportional to [\sin\theta]?

Torque equals \[pE\sin\theta\] because the effective lever arm is proportional to \[\sin\theta\] — zero when \[\theta=0^\circ\] (already aligned) and maximum when \[\theta=90^\circ\].

6. Can a dipole rotate indefinitely in a uniform field?

No. It will oscillate about equilibrium if undamped, or settle aligned if damping (friction) exists; energy considerations limit motion.

7. What causes simple harmonic motion for small displacements?

For small \[\theta\], \[\sin\theta \approx \theta\], so torque \[\tau \approx pE\theta\] gives linear restoring torque leading to SHM with \[\omega = \sqrt{pE/I}\].

8. If [\theta = 180^\circ], is equilibrium stable or unstable?

\[\theta=180^\circ\] (anti-parallel) is an unstable equilibrium — any small perturbation produces torque to move it away; energy is maximum there (\[U=+pE\]).

9. Does the magnitude of charges in the dipole matter individually for torque?

Only through the dipole moment \[p = q(2a)\]. Torque depends on \[p\], not separately on q or a.

10. If the field is reversed, what happens to torque direction?

Reversing \[\vec{E}\] reverses the sign of \[\vec{\tau} = \vec{p}\times\vec{E}\], so the torque direction reverses; dipole tends to align with new field direction.

11. How much work is done to rotate dipole from [0^\circ] to [180^\circ]?

Work = change in potential energy: \[ W = U(180^\circ) – U(0^\circ) = (+pE) – (-pE) = 2pE. \]

12. Why is torque zero when [\theta = 0^\circ]?

Because \[\sin 0^\circ = 0\], so \[\tau = pE\sin\theta = 0\]; dipole is already aligned (stable), no rotational tendency.

13. Can a dipole have translational acceleration if field is non-uniform?

Yes. In a non-uniform field the forces on +q and −q differ in magnitude, producing a net force and hence translation.

14. If dipole moment is zero, will torque be zero?

Yes. If \[p=0\] (e.g., q=0 or separation=0) there is no torque: \[\tau = pE\sin\theta = 0\].

15. Does the dipole experience torque in the absence of field?

No. Torque arises from interaction with \[\vec{E}\]; if \[\vec{E}=0\], \[\tau=0\].

7. FAQ / Common Misconceptions

1. Misconception: A dipole in a uniform field will always move towards higher potential.

No. In a uniform field net force = 0, so it won’t translate; it will only rotate until aligned.

2. Misconception: Torque depends on q and a separately.

Torque depends only on the dipole moment \[p = q(2a)\], so only the product matters.

3. Misconception: Potential energy is always positive.

No. \[U = -pE\cos\theta\] can be negative (minimum \[U=-pE\] at \[\theta=0\]).

4. Misconception: Reversing the dipole charges reverses p direction but leaves torque same.

Reversing charges reverses \[\vec{p}\]; since \[\vec{\tau}=\vec{p}\times\vec{E}\], torque direction reverses accordingly.

5. Misconception: Dipole oscillates forever even with damping.

Damping (friction, resistance) removes energy and oscillations decay; dipole settles aligned with the field.

6. Misconception: Work to rotate dipole depends on path.

Electric torque is conservative; work depends only on initial and final orientations (through U), not on path.

7. Misconception: Dipole in uniform field experiences force if its length is large.

No. Uniform field gives equal and opposite forces irrespective of length; net force remains zero (unless field varies across length).

8. Misconception: Maximum torque occurs at [\theta=90^\circ].

True — torque magnitude \[\tau=pE\sin\theta\] is maximum when \[\sin\theta=1\] (i.e., \[\theta=90^\circ\]).

9. Misconception: Potential energy zero at [\theta=0^\circ].

Only if zero reference chosen at that angle. Standard expression gives \[U(0)=-pE\]; choice of zero is arbitrary.

10. Misconception: Dipole moment changes in uniform field.

Intrinsic dipole moment depends on charge and separation; a rigid dipole’s p does not change due to external uniform field (unless charges move or polarisation occurs).

8. Practice Questions (With Step-By-Step Solutions)

Question 1 — Torque magnitude

A dipole of moment [p = 4.0\times10^{-7}\ \text{C·m}] is placed in a uniform electric field [E = 2.5\times10^{4}\ \text{N/C}]. Find the magnitude of torque when [\theta = 30^\circ].

Solution

Use [\tau = pE\sin\theta].
Substitute values:

[\tau] [= 4.0\times10^{-7}] [\times 2.5\times10^{4}] [\times \sin 30^\circ]

Compute (\sin 30^\circ = \dfrac{1}{2}).
Multiply numbers carefully:

First multiply [4.0\times10^{-7} \times 2.5\times10^{4}] [= (4.0 \times 2.5) \times 10^{-7 + 4}] [= 10.0 \times 10^{-3}] [= 1.0 \times 10^{-2}.]

Now multiply by (\sin 30^\circ = 0.5):

[\tau] [= 1.0\times10^{-2} \times 0.5] [= 0.5\times10^{-2}] [= 5.0\times10^{-3}\ \text{N·m}]

Answer: [5.0\times10^{-3}\ \text{N·m}].

Question 2 — Work done to rotate

Find the work required to rotate the same dipole ([p = 4.0\times10^{-7}\ \text{C·m}]) from [\theta_1 = 0^\circ] to [\theta_2 = 180^\circ] in the same field [E = 2.5\times10^{4}\ \text{N/C}].

Solution

Work = change in potential energy:

[W] [= U(180^\circ) – U(0^\circ)] [= (-pE\cos180^\circ) – (-pE\cos0^\circ)]

Evaluate cosines: [\cos180^\circ = -1,\ \cos0^\circ = 1].
Substitute:

[W] [= (-pE \times -1) – (-pE \times 1)] [= (pE) – (-pE)] [= 2pE]

Compute:

[2pE] [= 2 \times 4.0\times10^{-7} \times 2.5\times10^{4}]

Multiply numeric factors: [2 \times 4.0 \times 2.5 = 20.0.]
Powers of ten: [10^{-7} \times 10^{4} = 10^{-3}.]
So product: [20.0 \times 10^{-3} = 2.0 \times 10^{-2}.]
So:

[W = 2.0\times10^{-2}\ \text{J}]

Answer: [2.0\times10^{-2}\ \text{J}].

Question 3 — Small oscillation frequency

A small rigid dipole (moment of inertia about center [I = 8.0\times10^{-6}\ \text{kg·m}^2]) with dipole moment [p = 2.0\times10^{-6}\ \text{C·m}] is placed in uniform field [E = 1.0\times10^{3}\ \text{N/C}]. Find small oscillation angular frequency [\omega].

Solution

Use [\omega = \sqrt{\dfrac{pE}{I}}.]
Substitute:

[\omega] [= \sqrt{ \dfrac{2.0\times10^{-6} \times 1.0\times10^{3}}{8.0\times10^{-6}} }]

Compute numerator: [2.0\times10^{-6} \times 1.0\times10^{3}] [= 2.0 \times 10^{-3}.]
Divide by [8.0\times10^{-6}]:

[\dfrac{2.0\times10^{-3}}{8.0\times10^{-6}}] [= \dfrac{2.0}{8.0} \times 10^{-3 – (-6)}] [= 0.25 \times 10^{3}] [= 2.5\times10^{2}.]

Take square root:

[\omega] [= \sqrt{2.5\times10^{2}}] [= \sqrt{250} \approx 15.811…\ \text{rad/s}]

Round reasonably:

[\omega \approx 15.8\ \text{rad/s}]

Answer: [\omega \approx 15.8\ \text{rad/s}.]

Question 4 — Equilibrium orientation and stability

A dipole in uniform [\vec{E}] is initially at [\theta = 120^\circ]. Describe the torque direction and whether the dipole will move to stable or unstable equilibrium.

Solution

Torque magnitude: [\tau = pE\sin 120^\circ]. [(\sin120^\circ = \sin60^\circ)] [= \dfrac{\sqrt{3}}{2}.]
Direction of torque given by [\vec{\tau} = \vec{p} \times \vec{E}]. For [\theta=120^\circ] (between 90° and 180°), torque tends to reduce [\theta] (rotate toward 0°) — i.e., towards alignment.
Since [\theta=120^\circ] is nearer to unstable angle 180° but torque pushes it toward 0°, the dipole will rotate toward the stable equilibrium at [\theta=0^\circ].

Answer: Torque acts to decrease [\theta]; dipole moves toward stable equilibrium at [\theta=0^\circ].

Question 5 — Force in a slightly non-uniform field (qualitative)

A dipole with [p] is placed in a field that increases slightly with x (i.e., [E(x) = E_0 + \alpha x]). Will there be a net force? If yes, give expression to first order in [\alpha].

Solution (qualitative + first order expression)

In a non-uniform field, forces on +q and −q differ:

[F_+ = qE(x+a)], [\qquad] [F_- = -qE(x-a)]

Net force (approximate, Taylor-expand to first order in a):

[F_{\text{net}}] $\approx q[E(x)+aE'(x)]$ $- q[E(x)-aE'(x)]$ $= 2qa E'(x)$

Using [p = 2aq], we get:

[F_{\text{net}} \approx p,E'(x)]

For small (\alpha) where (E'(x)=\alpha):

[F_{\text{net}} \approx p\alpha]

Direction depends on sign of (\alpha) and orientation of dipole.

Answer: Yes. To first order, [F_{\text{net}} \approx p,E'(x)] (or [p\alpha] if (E’= \alpha)).