Dipole in uniform field

Rub a party balloon on your hair and hold it near a stream of water — the stream bends. The water molecules are tiny electric dipoles, and every dipole placed in an external electric field feels the field's push. But push how, exactly? Not as a net shove, it turns out — a uniform field pulls the dipole's two ends with equal and opposite force, so the dipole as a whole does not accelerate. What it does feel is a twist: a torque that tries to swing the dipole into line with the field, the same way a compass needle swings to point north. This article works out that torque from first principles, pins down where the twist vanishes (equilibrium) and where it is strongest, and solves two fully numerical torque problems — the same physics that explains why polar molecules like water orient themselves whenever an electric field is switched on.

The core insight

A dipole in a uniform electric field feels zero net force but a torque $\tau = pE\sin\theta$ that rotates it toward alignment with the field, vanishing (in stable or unstable equilibrium) at $\theta = 0°$ and $180°$ and maximal at $\theta = 90°$ .

The core idea

Picture a dipole of moment $\vec p = q(2a)$ sitting in a uniform external field $\vec E$ , with its axis making angle $\theta$ with the field.

Why the net force is zero. In a uniform field, $\vec E$ has the same value at both charges — one field vector, evaluated at two nearby points, gives the same answer. The positive charge feels $q\vec E$ , the negative charge feels $-q\vec E$ , and these are equal in size and opposite in direction, so they cancel exactly:

\vec F_{\text{net}} = q\vec E + (-q\vec E) = \vec 0

where: $q$ is the magnitude of each charge and $\vec E$ is the (single, shared) field vector. The dipole's centre of mass never accelerates in a uniform field, no matter the orientation $\theta$ .

Why there is still a torque. Zero net force does not mean nothing happens — the two equal and opposite forces act at different points, separated by the dipole length $2a$ , and any such pair forms a couple. A couple always produces a turning effect even though it produces no net push. The perpendicular distance between the two lines of action (the "moment arm" of the couple) is $2a\sin\theta$ , so the torque magnitude is force times arm:

\tau = (qE)(2a\sin\theta) = (q\cdot 2a)E\sin\theta = pE\sin\theta

where: $p = q(2a)$ is the dipole moment, $E$ is the field magnitude, and $\theta$ is the angle between $\vec p$ and $\vec E$ . In vector form, capturing both magnitude and the axis of rotation at once,

\vec\tau = \vec p \times \vec E

The direction of $\vec\tau$ , by the right-hand rule, is perpendicular to the plane containing $\vec p$ and $\vec E$ — it is an axial vector describing a rotation, not a push in any particular line.

Figure

A dipole tilted at angle theta to a uniform field showing equal and opposite forces on the two charges forming a couple that rotates the dipole toward alignment

The three landmark angles. At $\theta = 0°$ ( $\vec p$ parallel to $\vec E$ ), $\tau = pE\sin0° = 0$ — this is stable equilibrium, since nudging the dipole away from alignment produces a torque that restores it. At $\theta = 180°$ ( $\vec p$ anti-parallel to $\vec E$ ), $\tau = pE\sin180° = 0$ as well, but this is unstable equilibrium: the smallest disturbance produces a growing torque that carries the dipole all the way around to $\theta = 0°$ . At $\theta = 90°$ , $\tau = pE\sin90° = pE$ , the maximum torque the dipole can experience in that field.

Derivation

Derivation ATorque on a dipole in a uniform electric field

The goal is to show, starting only from the definition of dipole moment and the force on a point charge, that the turning effect of a uniform field on a dipole is $\tau = pE\sin\theta$ , and that this expression correctly predicts zero torque at alignment and anti-alignment and maximum torque at right angles — the three checkpoints every worked problem in this topic will test.

Step 1. Set up the geometry. Let the dipole have charges $+q$ and $-q$ separated by distance $2a$ , so the dipole moment has magnitude $p = q(2a)$ and points from $-q$ to $+q$ . Place the dipole in a uniform field $\vec E$ so that $\vec p$ makes angle $\theta$ with $\vec E$ .

Step 2. Write the two forces. Because the field is uniform, both charges sit in the same $\vec E$ . The force on $+q$ is $\vec F_+ = q\vec E$ and the force on $-q$ is $\vec F_- = -q\vec E$ . These are equal in magnitude and opposite in direction, so $\vec F_+ + \vec F_- = 0$ : no net force, confirming the dipole's centre of mass cannot accelerate under a uniform field at any $\theta$ .

Step 3. Find the moment arm of the couple. Since the two equal-and-opposite forces act along parallel lines separated by a perpendicular distance, they form a couple, and a couple's torque is (one force) times (perpendicular distance between the lines of action). Resolving the separation $2a$ into a component perpendicular to $\vec E$ gives $2a\sin\theta$ — this is the effective lever arm, since only the perpendicular component of the separation contributes to turning.

Step 4. Multiply force by arm. $\tau = (qE)(2a\sin\theta) = (q\cdot2a)E\sin\theta$ . Since $p = q\cdot2a$ by definition, this collapses to $\tau = pE\sin\theta$ , or in vector notation (which also fixes the rotation axis via the right-hand rule) $\vec\tau = \vec p\times\vec E$ .

Step 5. Check the checkpoints. At $\theta=0°$ : $\tau = pE\sin0° = 0$ — no twist, and since it is a minimum-energy alignment, any small tilt gets pulled back, so this is stable equilibrium. At $\theta=180°$ : $\tau = pE\sin180° = 0$ — also no twist, but here a small tilt gets pushed further away, so this is unstable equilibrium. At $\theta=90°$ : $\sin90°=1$ , so $\tau = pE$ , the largest value $\sin\theta$ can produce — the maximum torque. All three match physical intuition: a dipole already aligned or anti-aligned feels no twist, and a dipole crosswise to the field feels the strongest twist.

Worked example

A dipole of moment $p = 4\times10^{-8}\ \text{C}\,\cdot\,\text{m}$ sits in a uniform field of magnitude $E = 5\times10^{5}\ \text{N/C}$ , with its axis at $30°$ to the field. Find the torque, and state which direction it rotates the dipole.

Step 1. Identify the formula: $\tau = pE\sin\theta$ , with $p = 4\times10^{-8}\ \text{C}\,\cdot\,\text{m}$ , $E = 5\times10^{5}\ \text{N/C}$ , $\theta = 30°$ .

Step 2. Compute $pE$ first, to avoid a three-number multiplication in one line: $pE = (4\times10^{-8})\times(5\times10^{5}) = 2\times10^{-2}\ \text{N}\,\cdot\,\text{m} = 0.02\ \text{N}\,\cdot\,\text{m}$ .

Step 3. Multiply by $\sin30° = 0.5$ : $\tau = 0.02\times0.5 = 0.01\ \text{N}\,\cdot\,\text{m} = 1\times10^{-2}\ \text{N}\,\cdot\,\text{m}$ .

Step 4. State the direction: since $\theta=30°$ is neither $0°$ nor $180°$ , the torque is non-zero and acts to reduce $\theta$ — it rotates $\vec p$ toward alignment with $\vec E$ . The torque is $\tau = 1\times10^{-2}\ \text{N}\,\cdot\,\text{m}$ , directed so as to swing the dipole toward $\theta = 0°$ .

Competitive stretch

A dipole of charges $\pm3\ \mu\text{C}$ separated by $2\ \text{mm}$ sits with its axis perpendicular to a field of $2\times10^{4}\ \text{N/C}$ . Find the dipole moment and the torque, and say whether this is an equilibrium position.

Step 1. Dipole moment: $p = q(2a) = (3\times10^{-6})\times(2\times10^{-3}) = 6\times10^{-9}\ \text{C}\,\cdot\,\text{m}$ .

Step 2. With $\theta = 90°$ , $\sin\theta = 1$ , so $\tau = pE = (6\times10^{-9})\times(2\times10^{4}) = 1.2\times10^{-4}\ \text{N}\,\cdot\,\text{m}$ — the maximum torque possible for this dipole and field.

Step 3. Equilibrium check: equilibrium (stable or unstable) occurs only where $\tau = 0$ , i.e. $\theta = 0°$ or $180°$ . At $\theta=90°$ the torque is at its peak, not zero, so this is not an equilibrium position at all — the dipole will begin swinging toward $\theta=0°$ , the stable orientation.

Common traps

Assuming zero net force means the dipole is in equilibrium at every angle — force and torque are independent; a uniform field always gives zero force but a $\theta$ -dependent torque.
Mixing up stable ( $\theta=0°$ ) and unstable ( $\theta=180°$ ) equilibrium — both give $\tau=0$ , but only $\theta=0°$ has a restoring torque for a small displacement; at $\theta=180°$ a small displacement grows.
Dropping the $\sin\theta$ factor and writing $\tau=pE$ for any orientation — that value is only the maximum, reached at $\theta=90°$ ; the torque is zero at $0°$ and $180°$ .
Applying $\vec\tau=\vec p\times\vec E$ to a non-uniform field — there a net force also appears (the field differs at the two charges), which is outside this topic's uniform-field scope.

Ask the tutor: curious how this torque links to the dipole's potential energy $U=-\vec p\cdot\vec E$ later in the chapter, or want more torque-calculation practice?

If the net force on a dipole in a uniform field is always zero, how can it move at all?

It cannot translate, but it can rotate. Zero net force only rules out the centre of mass accelerating in a straight line; it says nothing about spin. The two equal and opposite forces act at different points along the dipole, forming a couple, and a couple produces pure rotation with no accompanying push. That is exactly why a dipole in a uniform field swings to align itself but never drifts sideways because of the field alone.

Why is theta equal to 180 degrees called unstable if the torque there is also zero, just like at theta equal to zero?

Zero torque only tells you the dipole is momentarily in equilibrium, not whether that equilibrium survives a nudge. At $\theta=0°$ , tilting the dipole slightly creates a small restoring torque, given by $\tau=pE\sin\theta\approx pE\theta$ for small $\theta$ , that pushes it back — stable. At $\theta=180°$ , the same small tilt creates a torque that grows and drives the dipole further away, eventually all the way around to $\theta=0°$ — unstable. The formula $\tau=pE\sin\theta$ is identical near both points; it is the direction the resulting torque points relative to the displacement that decides stability.

Does this uniform-field torque formula also work for a non-uniform field, like near a point charge?

No, and this is a common exam trap. In a non-uniform field the two charges of the dipole sit at points where $\vec E$ has different magnitude and/or direction, so the forces on them are no longer exactly equal and opposite. That leaves both a net force and a torque, and the simple $\tau=pE\sin\theta$ formula (derived assuming one shared $\vec E$ ) no longer applies as-is. This article's derivation, and the equilibrium classification, are valid strictly for a uniform external field.