PYQ Vault

MHT-CET Physics · Electrostatics

Electric Dipole — Field, Potential and Torque

Two equal and opposite charges a short distance apart form a dipole of moment p = q × 2a; far away its field falls as 1/r³, and in a uniform field it feels a torque but no net force.

All Electrostatics notesMHT-CET Physics strategy

Why this matters

9 PYQs, none HARD. Two things are asked: how the field and potential fall off along the axis and the equator, and the torque or work when a dipole turns in a uniform field. Every work question is the same subtraction, pE(cos θ₁ − cos θ₂).

Concept 1 of 2: Field and Potential on the Axis and the Equator

Far from a dipole its two charges almost cancel, so its field falls faster than a single charge's: as 1/r³, not 1/r². Along the axis the two contributions are in line and the field is twice as big as on the equatorial line at the same distance. On the equator the point is equidistant from +q and −q, so the potential there is exactly zero.

Definition

  • Dipole moment p=q×2ap = q \times 2a, directed from −q-q to +q+q.
  • Axis: Ea=2kpr3E_a = \dfrac{2kp}{r^3}, along p⃗\vec p; Va=kpr2V_a = \dfrac{kp}{r^2}.
  • Equator: Ee=kpr3E_e = \dfrac{kp}{r^3}, opposite to p⃗\vec p; Ve=0V_e = 0.
  • So Ea=2EeE_a = 2E_e at the same distance, field ∝1r3\propto \dfrac{1}{r^3}, potential ∝1r2\propto \dfrac{1}{r^2}.
  • Two dipoles on one line: where their axial fields cancel, p1x3=p2(d−x)3\dfrac{p_1}{x^3} = \dfrac{p_2}{(d - x)^3}.

Short dipole

Eaxis=2kpr3,Eequator=kpr3,Vaxis=kpr2E_{\text{axis}} = \frac{2kp}{r^3}, \quad E_{\text{equator}} = \frac{kp}{r^3}, \quad V_{\text{axis}} = \frac{kp}{r^2}

Worked example

A short dipole has p=4×10−9p = 4 \times 10^{-9} C m. Field 20 cm from it on the axis, and on the equator?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q23Easy

Example 1 · Electrostatics · Electric Dipole — Field, Potential, Torque

If EaE_a and EqE_q represent the electric field intensity due to a short dipole at a point on its axial line and on the equatorial line at the same distance rr from the centre of the dipole, then

Field power versus potential power

Field falls as 1/r31/r^3, potential as 1/r21/r^2. A question asking 'the potential on the axis is proportional to' has 1/r31/r^3 as the planted wrong answer.
Drill 2 more on field and potential on the axis and the equator

Concept 2 of 2: Torque, Energy and Work for a Dipole in a Uniform Field

In a uniform field the forces on +q and −q are equal and opposite, so the dipole does not move off but it turns, trying to line up with the field. Lined up, its energy is lowest (−pE); turned right round it is highest (+pE). The work to turn it is the rise in that energy.

Definition

  • Torque τ=pEsin⁡θ\tau = pE\sin\theta, largest (pEpE) at θ=90∘\theta = 90^\circ. Net force in a uniform field: zero.
  • Energy U=−pEcos⁡θU = -pE\cos\theta: minimum −pE-pE at θ=0\theta = 0 (stable), maximum +pE+pE at 180∘180^\circ (unstable).
  • Work to turn from θ1\theta_1 to θ2\theta_2: W=pE(cos⁡θ1−cos⁡θ2)W = pE(\cos\theta_1 - \cos\theta_2). From aligned: 90∘90^\circ costs pEpE, 60∘60^\circ costs pE2\frac{pE}{2}, 180∘180^\circ costs 2pE2pE.
  • Length from torque: 2a=τmax⁡qE2a = \dfrac{\tau_{\max}}{qE}.

Dipole in a uniform field

τ=pEsin⁡θ,U=−pEcos⁡θ,Wθ1→θ2=pE(cos⁡θ1−cos⁡θ2)\tau = pE\sin\theta, \qquad U = -pE\cos\theta, \qquad W_{\theta_1 \to \theta_2} = pE(\cos\theta_1 - \cos\theta_2)

Worked example

A dipole of charges ±3 μ\pm 3\,\muC, 2 cm apart, is in a field of 5×1045 \times 10^4 N/C. Find the maximum torque and the work to turn it from aligned to reversed.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q18Moderate

Example 2 · Electrostatics · Electric Dipole — Field, Potential, Torque

The work done in rotating a dipole placed parallel to the electric field through 180∘180^\circ is 'ww'. So the work done in rotating it through 60∘60^\circ is (cos⁡0∘=1,cos⁡60∘=12,cos⁡180∘=−1\cos 0^\circ=1,\cos 60^\circ=\frac{1}{2},\cos 180^\circ=-1)

Using sin θ for the work

Torque uses sin⁡θ\sin\theta; work and energy use cos⁡θ\cos\theta. Turning an aligned dipole through 60∘60^\circ costs pE(1−cos⁡60∘)=pE2pE(1 - \cos 60^\circ) = \frac{pE}{2}, not pEsin⁡60∘pE\sin 60^\circ.
Drill 5 more on torque, energy and work for a dipole in a uniform field

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Field and Potential on the Axis and the Equator

    Short dipole

    Eaxis=2kpr3,Eequator=kpr3,Vaxis=kpr2E_{\text{axis}} = \frac{2kp}{r^3}, \quad E_{\text{equator}} = \frac{kp}{r^3}, \quad V_{\text{axis}} = \frac{kp}{r^2}
  • Torque, Energy and Work for a Dipole in a Uniform Field

    Dipole in a uniform field

    τ=pEsin⁡θ,U=−pEcos⁡θ,Wθ1→θ2=pE(cos⁡θ1−cos⁡θ2)\tau = pE\sin\theta, \qquad U = -pE\cos\theta, \qquad W_{\theta_1 \to \theta_2} = pE(\cos\theta_1 - \cos\theta_2)

Watch out for (2)

Test yourself on Electrostatics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Electrostatics chapter testDrill all 9 questions

Related notes