Electricity-And-Magnetism · Unit 3 · Video 5 · Interactive Practice

Torque, Energy and Force on a Dipole in a Field

IKey Formulas

FormulaNameWhen it applies
F⃗net=qE⃗−qE⃗=0\vec{F}_{\text{net}} = q\vec{E} - q\vec{E} = 0Net forceUniform field
τ⃗=p⃗×E⃗\vec{\tau} = \vec{p}\times\vec{E},  ∣τ⃗∣=pEsin⁡θ\ |\vec{\tau}| = pE\sin\thetaTorque (turns p⃗\vec{p} toward E⃗\vec{E})Uniform field, θ\theta from E⃗\vec{E} to p⃗\vec{p}
U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p}\cdot\vec{E} = -pE\cos\thetaPotential energyZero chosen at θ=π/2\theta = \pi/2
F⃗e=p dEdx i^\vec{F}_e = p\,\dfrac{dE}{dx}\,\hat{i}Net forceSmall dipole along xx in a non-uniform field

Key Insight: A uniform field can only rotate a dipole; moving it requires a field that changes from place to place.

IIVisualization 1 — The Couple in a Uniform Field

The forces ±qE⃗\pm q\vec{E} cancel, but their lines of action are offset by 2asin⁡θ2a\sin\theta, so they turn p⃗\vec{p} toward E⃗\vec{E}.

IIIVisualization 2 — Work and the Energy Curve

Turning from θ0\theta_0 to θ\theta, the field does work W=pE(cos⁡θ−cos⁡θ0)W = pE(\cos\theta - \cos\theta_0), and U=−pEcos⁡θU = -pE\cos\theta falls by exactly WW.

IVVisualization 3 — Net Force in a Non-Uniform Field

When EE differs at the two ends, the end forces no longer cancel, leaving F⃗e=p (dE/dx) i^\vec{F}_e = p\,(dE/dx)\,\hat{i}.

💡 A charged comb induces a dipole moment in a neutral scrap of paper along the comb's field, so by this result the scrap is pulled toward the stronger field near the comb, whatever the sign of the comb's charge.

VQuiz Questions

Problem 1 · Torque Magnitude

Given: Charges ±3.0 nC\pm 3.0\ \text{nC} separated by 2a=4.0 mm2a = 4.0\ \text{mm} sit in a uniform field E=5.0×104 N/CE = 5.0\times10^{4}\ \text{N/C}, with p⃗\vec{p} at θ=30°\theta = 30\degree to E⃗\vec{E}. Find ∣τ⃗∣|\vec{\tau}|.

✅ Correct! p=q(2a)=1.2×10−11 C⋅mp = q(2a) = 1.2\times10^{-11}\ \text{C}\cdot\text{m}, and pEsin⁡30°=3.0×10−7 N⋅mpE\sin 30\degree = 3.0\times10^{-7}\ \text{N}\cdot\text{m}.
❌ Not quite. The dipole moment is p=2qap = 2qa, using the full separation 2a=4.0 mm2a = 4.0\ \text{mm}, not the half-length aa.
❌ Not quite. The torque goes as sin⁡θ\sin\theta: it vanishes when p⃗∥E⃗\vec{p} \parallel \vec{E}. You used cos⁡θ\cos\theta.
❌ Not quite. pEpE is the largest torque, reached only at θ=90°\theta = 90\degree. Include the factor sin⁡30°=12\sin 30\degree = \tfrac{1}{2}.
❌ Not quite. Compute p=2qap = 2qa first, then ∣τ⃗∣=pEsin⁡θ|\vec{\tau}| = pE\sin\theta.
Show solution

The dipole moment uses the full separation:

p=q(2a)=(3.0×10−9 C)(4.0×10−3 m)=1.2×10−11 C⋅mp = q(2a) = (3.0\times10^{-9}\ \text{C})(4.0\times10^{-3}\ \text{m}) = 1.2\times10^{-11}\ \text{C}\cdot\text{m}

The torque magnitude is

∣τ⃗∣=pEsin⁡θ=(1.2×10−11)(5.0×104)sin⁡30°=(6.0×10−7)(0.5)=3.0×10−7 N⋅m|\vec{\tau}| = pE\sin\theta = (1.2\times10^{-11})(5.0\times10^{4})\sin 30\degree = (6.0\times10^{-7})(0.5) = 3.0\times10^{-7}\ \text{N}\cdot\text{m}

Its direction, p⃗×E⃗\vec{p}\times\vec{E}, turns p⃗\vec{p} toward E⃗\vec{E}.

Problem 2 · Where Is U Lowest?

Given: A dipole with p=2.0×10−11 C⋅mp = 2.0\times10^{-11}\ \text{C}\cdot\text{m} in a uniform field E=3.0×104 N/CE = 3.0\times10^{4}\ \text{N/C}, with U(π/2)=0U(\pi/2) = 0. Find the orientation of lowest potential energy and UU there.

✅ Correct! U=−pEcos⁡θU = -pE\cos\theta is most negative at θ=0\theta = 0: Umin⁡=−pE=−6.0×10−7 JU_{\min} = -pE = -6.0\times10^{-7}\ \text{J}, the stable equilibrium.
❌ Not quite. U=0U = 0 at π/2\pi/2 is only the chosen reference. Negative energies lie below it, and π/2\pi/2 is where the torque is largest, not an equilibrium.
❌ Check the sign. U=−p⃗⋅E⃗U = -\vec{p}\cdot\vec{E}. At θ=π\theta = \pi, cos⁡π=−1\cos\pi = -1 gives U=+pEU = +pE, the maximum.
❌ Not quite. Both are equilibria, but U(0)=−pEU(0) = -pE while U(π)=+pEU(\pi) = +pE. Only θ=0\theta = 0 is the minimum (stable).
Show solution

With the zero at θ=π/2\theta = \pi/2, U(θ)=−pEcos⁡θU(\theta) = -pE\cos\theta, and pE=(2.0×10−11)(3.0×104)=6.0×10−7 JpE = (2.0\times10^{-11})(3.0\times10^{4}) = 6.0\times10^{-7}\ \text{J}.

U(0)=−pE=−6.0×10−7 J,U(π)=+pE=+6.0×10−7 JU(0) = -pE = -6.0\times10^{-7}\ \text{J},\qquad U(\pi) = +pE = +6.0\times10^{-7}\ \text{J}

The torque pEsin⁡θpE\sin\theta vanishes at both ends, but a small tilt from θ=0\theta = 0 produces a torque back toward E⃗\vec{E} (stable, minimum UU), while a tilt from θ=π\theta = \pi turns the dipole further away (unstable, maximum UU).

Problem 3 · Work to Flip a Dipole

Given: The same dipole (pE=6.0×10−7 JpE = 6.0\times10^{-7}\ \text{J}) turns from θ0=60°\theta_0 = 60\degree to θ=180°\theta = 180\degree. Find the work WW done by the field and the change ΔU\Delta U.

Work done by the field

Change in potential energy

✅ Correct! Turning away from E⃗\vec{E}, the field does negative work, W=−1.5 pEW = -1.5\,pE, and UU rises by 1.5 pE1.5\,pE.
❌ Check the work. W=pE(cos⁡θ−cos⁡θ0)W = pE(\cos\theta - \cos\theta_0) with the final angle first: cos⁡180°=−1\cos 180\degree = -1 and cos⁡60°=12\cos 60\degree = \tfrac{1}{2}.
❌ Check ΔU\Delta U. The change in potential energy is the negative of the field's work: ΔU=−W\Delta U = -W.
Show solution

Integrate dW=−pEsin⁡θ′ dθ′dW = -pE\sin\theta'\,d\theta' from θ0\theta_0 to θ\theta:

W=pE(cos⁡θ−cos⁡θ0)=(6.0×10−7)(cos⁡180°−cos⁡60°)=(6.0×10−7)(−1−0.5)=−9.0×10−7 JW = pE(\cos\theta - \cos\theta_0) = (6.0\times10^{-7})\big(\cos 180\degree - \cos 60\degree\big) = (6.0\times10^{-7})(-1 - 0.5) = -9.0\times10^{-7}\ \text{J} ΔU=−W=+9.0×10−7 J\Delta U = -W = +9.0\times10^{-7}\ \text{J}

Check with U=−pEcos⁡θU = -pE\cos\theta: U0=−3.0×10−7 JU_0 = -3.0\times10^{-7}\ \text{J}, U=+6.0×10−7 JU = +6.0\times10^{-7}\ \text{J}, so ΔU=+9.0×10−7 J\Delta U = +9.0\times10^{-7}\ \text{J}. The dipole ended farther from the field direction, so W<0W < 0.

Problem 4 · Dipole in a Weakening Field

Given: Charges ±2.0 nC\pm 2.0\ \text{nC} separated by 2a=2.0 mm2a = 2.0\ \text{mm} lie on the xx axis with p⃗\vec{p} along +i^+\hat{i}. The field there is E⃗=E(x) i^\vec{E} = E(x)\,\hat{i} with E>0E > 0 and dE/dx=−2.0×106 N/(C⋅m)dE/dx = -2.0\times10^{6}\ \text{N}/(\text{C}\cdot\text{m}). Find the net force on the dipole.

✅ Correct! F⃗e=p (dE/dx) i^=−8.0 μN i^\vec{F}_e = p\,(dE/dx)\,\hat{i} = -8.0\ \mu\text{N}\,\hat{i}: with p⃗\vec{p} along E⃗\vec{E}, the dipole moves toward the stronger field at smaller xx.
❌ Not quite. Zero net charge gives zero net force only in a uniform field. Here the field at the two ends differs.
❌ Check the sign. dE/dx<0dE/dx < 0, so p dE/dxp\,dE/dx is negative: the force on the −q-q end, where the field is stronger, wins.
❌ Not quite. The dipole moment is p=q(2a)=4.0×10−12 C⋅mp = q(2a) = 4.0\times10^{-12}\ \text{C}\cdot\text{m}, not qaqa.
Show solution
p=q(2a)=(2.0×10−9 C)(2.0×10−3 m)=4.0×10−12 C⋅mp = q(2a) = (2.0\times10^{-9}\ \text{C})(2.0\times10^{-3}\ \text{m}) = 4.0\times10^{-12}\ \text{C}\cdot\text{m} F⃗e=p dEdx i^=(4.0×10−12)(−2.0×106) i^=−8.0×10−6 N i^\vec{F}_e = p\,\frac{dE}{dx}\,\hat{i} = (4.0\times10^{-12})(-2.0\times10^{6})\,\hat{i} = -8.0\times10^{-6}\ \text{N}\ \hat{i}

The −q-q end sits at smaller xx, where the field is stronger, so its leftward force qE(x−a)qE(x - a) exceeds the rightward force qE(x+a)qE(x + a) on +q+q. The net force is 8.0 μN8.0\ \mu\text{N} toward −x-x, the stronger field.

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