Dipole in a Uniform Field: Torque & Energy

Your microwave oven runs equation (1) a few billion times per second. The magnetron fills the cavity with a field oscillating at $2.45\ \mathrm{GHz}$; each reversal torques the water dipoles in the food, their twisting rubs against neighbouring molecules, and the rotational work degrades into heat. One honest correction to a widespread myth: $2.45\ \mathrm{GHz}$ is not a special resonance of the water molecule — it is an engineering compromise chosen so the waves penetrate centimetres into food instead of burning only the surface. The physics is not resonance but relentless torque-and-friction, which is why oil (non-polar, tiny $p$) heats poorly in a microwave while water-rich food heats fast. The same torque runs the screen you may be reading this on. Each pixel of a liquid-crystal display holds rod-like polar molecules; a small voltage across the pixel applies a field, the field torques the molecular dipoles into a new orientation, and the reoriented layer rotates the polarisation of light passing through — switching the pixel between bright and dark. Billions of pixels, each a dipole-alignment experiment from this module, refreshed sixty times a second.

Take the dipole from the last module — charges $+q$ and $-q$ glued a distance $d$ apart, dipole moment $p = qd$ pointing from $-q$ to $+q$ — and lower it into a uniform field $\vec{E}$, the kind that lives between large parallel plates. Each end now feels its own force. The $+q$ end is pushed along the field with force $qE$; the $-q$ end is pushed against the field with force $qE$. Same magnitude, opposite directions. Add them: zero. The dipole as a whole is not pushed anywhere. This is worth saying loudly, because it surprises people — a neutral object with zero net charge feels zero net force in a uniform field, no matter how strong the field is. A dipole between capacitor plates does not drift toward either plate. But watch what the two forces do together. They act at different points — one at each end of the dipole — and they point in opposite directions. Two equal, opposite, parallel forces acting at separated points are what mechanics calls a couple: they cannot translate the object, but they rotate it. The $+q$ end gets pulled one way, the $-q$ end the other, and the dipole swings like a compass needle until $\vec{p}$ lines up with $\vec{E}$. That is the entire phenomenon: a uniform field grips a dipole purely through rotation.

How hard is the twist? A couple's torque is one force times the perpendicular distance between the two lines of action. Tilt the dipole at angle $\theta$ to the field and that perpendicular separation is $d\sin\theta$ — the geometry does the rest, giving $\tau = qE \cdot d\sin\theta = pE\sin\theta$. At $\theta = 90°$, the arm is the full length $d$ and the torque peaks. At $\theta = 0°$, the two forces pull head-to-tail along the same line: no arm, no torque, the dipole rests. And at $\theta = 180°$ the torque is also zero — but that resting point hides a trap we will spring shortly. The compass analogy is exact and worth keeping. A compass needle is a magnetic dipole in the Earth's magnetic field; our electric dipole in $\vec{E}$ behaves identically — swing it off alignment and it fights back hardest at $90°$, not at all when aligned or exactly reversed. When you meet current loops behaving as magnetic dipoles later in the course, the formulas will be $\tau = mB\sin\theta$ and $U = -mB\cos\theta$: the same physics wearing a different costume.

Here is the headline result, stated cleanly. For a dipole of moment $p$ at angle $\theta$ to a uniform field $E$, the magnitude of the torque is

where: $p = qd$ — the dipole moment in $\mathrm{C\,m}$; $E$ — the field strength in $\mathrm{N/C}$ (equivalently $\mathrm{V/m}$); $\theta$ — the angle between $\vec{p}$ and $\vec{E}$; $\tau$ — in $\mathrm{N\,m}$. Read it physically: the torque is proportional to how strong the dipole is, how strong the field is, and how far out of line they are. The $\sin\theta$ means the torque always acts to reduce $\theta$ — it is a restoring twist toward alignment, never away from it. In vector language, which CBSE requires you to state and JEE loves to test with direction questions,

where: the cross product bundles magnitude and direction in one line. The magnitude is $pE\sin\theta$ automatically, and the direction follows the right-hand rule: curl the fingers of your right hand from $\vec{p}$ toward $\vec{E}$ through the smaller angle, and your thumb gives $\vec{\tau}$ — perpendicular to both, along the axis about which the dipole rotates. If $\vec{p}$ and $\vec{E}$ both lie on this page, the torque points into or out of the page. You met this same cross-product machinery in rotational mechanics ($\vec{\tau} = \vec{r}\times\vec{F}$); this is it, earning its keep in electrostatics.

A practical anchor before moving on. A water molecule has $p = 6.2\times10^{-30}\ \mathrm{C\,m}$. In a strong laboratory field of $10^6\ \mathrm{N/C}$, its maximum torque is $pE = 6.2\times10^{-24}\ \mathrm{N\,m}$ — invisibly small on our scale, yet acting on something as light as a molecule it drives rotations billions of times per second. The formulas of this module run from exam problems with $10^{-4}\ \mathrm{N\,m}$ torques down to the machinery of your microwave without changing a symbol.

Since the field twists the dipole, holding the dipole at an angle takes effort, and rotating it against the field stores energy — exactly like winding a spring. The potential energy of a dipole at angle $\theta$, with the standard choice $U = 0$ at $\theta = 90°$, is

where: the minus sign and the cosine carry all the physics. At $\theta = 0°$ (aligned): $U = -pE$, the lowest possible — the bottom of the energy valley. At $\theta = 90°$: $U = 0$, the chosen reference. At $\theta = 180°$ (anti-aligned): $U = +pE$, the highest — the summit. In compact vector form, $U = -\vec{p}\cdot\vec{E}$. The dot product here and the cross product in equation (2) make a tidy pair: torque cares about the misalignment ($\sin\theta$), energy about the alignment ($\cos\theta$). For work-done questions — the JEE and NEET favourite — subtract two values of $U$. The work an external agent must do to rotate the dipole slowly from $\theta_1$ to $\theta_2$ equals the energy gained:

where: the formula handles every case without special memorisation. From $0°$ to $180°$ (the full flip): $W = pE(1-(-1)) = 2pE$. From $0°$ to $90°$: $W = pE$. From $90°$ to $180°$: $W = pE$ again — each quarter-turn away from alignment costs the same $pE$. The work done by the field is the negative of $W_{\text{ext}}$; questions alternate between the two, and reading which one is asked is worth a mark every time.

The landscape settles the equilibrium question cleanly. Both $\theta = 0°$ and $\theta = 180°$ have zero torque — both are equilibria. But nudge the dipole slightly at each. At $0°$, any small rotation raises $U$, and the torque that appears pushes it back: stable. At $180°$, any small rotation lowers $U$, and the torque that appears pushes it further away: unstable — the dipole swings all the way around to alignment, converting $2pE$ of stored energy into rotational motion. Exam questions that ask "for what orientation is the dipole in equilibrium?" deserve both angles with their stability labels; naming only $0°$ drops half the marks. One more result worth carrying, at stating level. A dipole disturbed slightly from stable equilibrium oscillates — the restoring torque $\tau \approx -pE\theta$ for small angles makes it a rotational cousin of the pendulum, with time period $T = 2\pi\sqrt{I/pE}$, where $I$ is the dipole's moment of inertia about the rotation axis. JEE Advanced has asked exactly this; recognise the pattern $T = 2\pi\sqrt{\text{inertia}/\text{stiffness}}$ from your oscillations chapter and quote it.

Board level — torque and energy on a tilted dipole (CBSE/ISC pattern)

A dipole consists of charges $\pm 1.0\ \mu\mathrm{C}$ separated by $2.0\ \mathrm{cm}$, placed at $30°$ to a uniform field $E = 5.0\times10^{4}\ \mathrm{N/C}$. Find (a) the dipole moment, (b) the torque on the dipole, (c) its potential energy, and (d) the work an external agent does to rotate it from $30°$ to $90°$.

Step 1. Dipole moment: $p = qd = (1.0\times10^{-6})(2.0\times10^{-2}) = 2.0\times10^{-8}\ \mathrm{C\,m}$, directed from $-q$ to $+q$. (State the direction — it carries a mark.)

Step 2. Torque from equation (1): $\tau = pE\sin 30° = (2.0\times10^{-8})(5.0\times10^{4})(0.5) = \boxed{5.0\times10^{-4}\ \mathrm{N\,m}}$, acting to rotate $\vec{p}$ toward $\vec{E}$.

Step 3. Potential energy from equation (3): $U = -pE\cos 30° = -(2.0\times10^{-8})(5.0\times10^{4})(0.866) = \boxed{-8.66\times10^{-4}\ \mathrm{J}}$. Negative — the dipole is on the valley side of the reference, closer to alignment than to $90°$.

Step 4. Work to rotate $30° \to 90°$, from equation (4): $W = pE(\cos 30° - \cos 90°) = (10^{-3})(0.866 - 0) = \boxed{+8.66\times10^{-4}\ \mathrm{J}}$ — exactly $-U$ from Step 3, as it must be, since the target orientation is the $U = 0$ reference.

Step 5. The marks-scheme sentence: "The net force on the dipole is zero because the field is uniform; the field exerts only a torque." Writing that one line pre-empts the standard follow-up and shows the examiner you know why translation never entered the problem.

Competitive level — rotate, release, and count the energy (JEE-Main/NEET pattern)

A dipole with $p = 4.0\times10^{-9}\ \mathrm{C\,m}$ sits aligned with a uniform field $E = 3.0\times10^{4}\ \mathrm{N/C}$. (a) How much work is needed to rotate it to $60°$? (b) What is the maximum torque it experiences during the rotation? (c) If released from $60°$, what is its kinetic energy when it swings back through alignment?

Step 1. Compute the one product everything uses: $pE = (4.0\times10^{-9})(3.0\times10^{4}) = 1.2\times10^{-4}\ \mathrm{J}$. (Do this first; all three answers are multiples of it.)

Step 2. Work from $0°$ to $60°$: $W = pE(\cos 0° - \cos 60°) = (1.2\times10^{-4})(1 - 0.5) = \boxed{6.0\times10^{-5}\ \mathrm{J}}$.

Step 3. Maximum torque during a $0° \to 60°$ rotation occurs at the largest angle reached, $60°$ (torque grows with $\theta$ up to $90°$): $\tau_{\max} = pE\sin 60° = (1.2\times10^{-4})(0.866) = \boxed{1.04\times10^{-4}\ \mathrm{N\,m}}$. Distractor alert: $pE = 1.2\times10^{-4}\ \mathrm{N\,m}$ (the $90°$ value) will be among the options — the dipole never reaches $90°$ here.

Step 4. Released from $60°$: energy conservation between orientations. $KE = U(60°) - U(0°) = pE(1 - \cos 60°) = \boxed{6.0\times10^{-5}\ \mathrm{J}}$ — the same number as Step 2, because the field returns exactly the work the agent invested. Timing note: 90 seconds. Compute $pE$ once, then every part is one multiplication. The only thinking point is Step 3's "maximum torque within the range" — five seconds of care that defeats the planted $90°$ option.

Sine–cosine swap. The single biggest mark-killer here: $\tau = pE\sin\theta$, $U = -pE\cos\theta$. Anchor it physically — torque must vanish when aligned ($\sin 0 = 0$ ✓), energy must be lowest when aligned ($-\cos$ gives $-pE$ ✓). Two seconds of checking beats re-memorising.

Which angle did they give you? Some questions quote the angle between the dipole and the plates, or from the perpendicular to the field. The formulas need $\theta$ measured from $\vec{E}$ itself. If the dipole makes $30°$ with a plate, it makes $60°$ with the field between the plates — convert before substituting.

Work by the field vs work by the agent. They differ by a sign: $W_{\text{ext}} = U_2 - U_1$, $W_{\text{field}} = -(U_2 - U_1)$. NEET options routinely include both magnitudes with both signs. Read "work done by the external agent" vs "work done by the field" twice.

The missing minus in U. Writing $U = pE\cos\theta$ inverts the entire landscape — it makes $180°$ look stable. If your stability conclusions feel backwards, hunt for this sign first.

Flip costs 2pE, not pE. From aligned to anti-aligned is $\cos 0° - \cos 180° = 2$. Quoting $pE$ for the flip (or $2pE$ for a $90°$ rotation) is a classic one-mark leak.

"Neutral, so nothing happens." Zero net charge kills the net force in a uniform field, not the torque. The dipole rotates. And the moment the field is non-uniform, even the force returns — that case is flagged, not tested, at this level.

Equilibrium without stability labels. "$\theta = 0°$ and $180°$" earns partial credit at best. Attach stable and unstable with the one-line energy justification.

Constants and conventions: boards accept $k = 9\times10^{9}\ \mathrm{N\,m^2/C^2}$ where it appears; keep $p$ in $\mathrm{C\,m}$ (convert $\mu\mathrm{C}$ and cm first — a $10^{-8}$ slip in $p$ poisons every later line).

Texts. NCERT Physics Part 1, Ch. 1 (Electric Charges and Fields, §1.11 — dipole in a uniform external field) — the CBSE/NEET canonical treatment · H. C. Verma, Concepts of Physics Vol. 2, Ch. 29 · Halliday, Resnick & Walker, Ch. 22 (torque and energy of a dipole).

Provenance. Standard results per NCERT Ch. 1 and HRW Ch. 22. A verifier should recompute: board example — $p = 2.0\times10^{-8}\ \mathrm{C\,m}$, $\tau = 5.0\times10^{-4}\ \mathrm{N\,m}$, $U = -8.66\times10^{-4}\ \mathrm{J}$, $W(30°\to90°) = +8.66\times10^{-4}\ \mathrm{J}$; competitive example — $pE = 1.2\times10^{-4}\ \mathrm{J}$, $W(0°\to60°) = 6.0\times10^{-5}\ \mathrm{J}$, $\tau(60°) = 1.04\times10^{-4}\ \mathrm{N\,m}$, $KE = 6.0\times10^{-5}\ \mathrm{J}$; water anchor $pE = 6.2\times10^{-24}\ \mathrm{N\,m}$ at $10^{6}\ \mathrm{N/C}$.

Prerequisites. The Electric Dipole & Dipole Moment · Torque (rotational mechanics). Leads to. Potential of a Dipole · Current Loops as Magnetic Dipoles (the magnetic twin) · Torque & Force on a Dipole in a Field (UG deep dive, for the non-uniform-field force).