Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E, as shown in Fig. 2.16. As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque t t t t t given by t = t = t = t = t = p × E (2.30) which will tend to rotate it (unless p is parallel or antiparallel to E). Suppose an external torque text is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle q0 to angle q1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by t ( ) cos cos pE θ θ 0 1 = − (2.31)
-- NCERT Class 12 Physics, Ch. 2, p. 60Electrical Potential Energy Dipole Field
8 MCQs9-step worked example
Source: NCERT ElectrostaticsOfficial key: NTA-verifiedLast updated: 10 Oct 2026
Try this first
In a uniform electric field E, the potential energy of a dipole of moment p at angle θ to the field, with the zero of energy taken at θ = 90°, is:
- A.U = pE sin θ
- B.U = pE cos θ
- C.U = −pE cos θ
- D.U = −pE sin θ
Tap to see the answer
Answer: C. Based on NCERT Class 12 Physics, Chapter 2, pages 60–61: the work stored as potential energy, with θ₀ = 90°, gives U(θ) = −pE cos θ.
A is wrong: A is wrong because pE sin θ has the form of the torque magnitude τ = pE sin θ, not of the energy.
B is wrong: B is wrong because it has the right trigonometric function but the wrong sign; with U = 0 at 90° the aligned dipole lies below zero.
D is wrong: D is wrong because it is a sine form, which belongs to torque; the energy involves the cosine of the angle.
Electrical Potential Energy Dipole Field, explained for NEET
The trap: using the torque expression where the energy expression belongs. The torque on a dipole is τ = pE sin θ and is zero when the dipole is aligned; the potential energy is U = −pE cos θ and is at its lowest, −pE, when the dipole is aligned. A student who plugs sin θ into an energy question gets the wrong number, and a student who thinks an aligned dipole has zero energy gets the wrong sign.
What the field does. NCERT Class 12 Physics, Chapter 2, page 60: in a uniform field a dipole feels no net force, but it feels a torque τ = p × E that tends to rotate it, unless p is parallel or antiparallel to E.
Where U comes from. Page 60 imagines an external torque that just neutralises this torque and turns the dipole very slowly from angle θ₀ to θ₁. The work it does is
W = pE (cos θ₀ − cos θ₁)
and this work is stored as the potential energy of the system (page 60). As with any potential energy, the zero is a free choice. NCERT takes θ₀ = 90° (pages 60–61), which gives
U(θ) = −pE cos θ
Why 90°. With this choice the work done in bringing +q and −q into place is equal and opposite, so the two cancel (page 61). U is then 0 at 90°, −pE at 0° (aligned) and +pE at 180° (opposed).
Only differences matter. The change U(θ₁) − U(θ₀) = pE (cos θ₀ − cos θ₁) does not depend on the zero you chose (it follows from the formula above). NCERT's worked example on page 61 shows a loss in potential energy as a dipole turns toward the field, released as heat.
Bridge to NEET. Questions give p, E and two angles and ask for U, ΔU or the external work. Evaluate each cosine, subtract in the right order, and attach the sign.
Watch out: the work done by the external torque equals the change in U, pE (cos θ₀ − cos θ₁), with θ₀ the starting angle. Swap the angles and you flip the sign; the work done by the field itself is the negative of this.
How do you solve a Electrical Potential Energy Dipole Field question? A worked example
- 1
Given
A dipole of moment 6.0 × 10⁻⁶ C m lies at 120° to a uniform electric field of 2.0 × 10⁵ N/C. An external torque turns it very slowly to 45° to the field.
- 2
Required
The work done by the external torque.
- 3
Concept
The external torque only balances the field's torque, so its work is stored as potential energy: W = pE (cos θ₀ − cos θ₁), which equals U(θ₁) − U(θ₀) with U = −pE cos θ (NCERT Class 12 Physics, Chapter 2, pages 60–61).
- 4
Formula
W = pE (cos θ₀ − cos θ₁), with pE = (6.0 × 10⁻⁶)(2.0 × 10⁵).
- 5
Substitution
pE = 1.2 J, θ₀ = 120°, θ₁ = 45°, so W = 1.2 × (cos 120° − cos 45°).
- 6
Calculation
cos 120° = −0.50 and cos 45° = 0.707, so W = 1.2 × (−0.50 − 0.707) = 1.2 × (−1.207) = −1.4 J.
The angles are exact values and do not contribute to the significant-figure count; the two measured inputs carry two significant figures each. - 7
Final answer
W = −1.4 J. The work done by the external torque is negative: the field itself turns the dipole toward alignment, so U falls by 1.4 J.
- 8
Common trap
Writing the energy as pE sin θ. That gives 1.2 × (sin 45° − sin 120°) = 1.2 × (0.707 − 0.866) = −0.19 J, which is far from the correct −1.4 J. The torque uses sin θ; the energy uses cos θ.
- 9
Similar NEET-style question
A dipole with p = 5.0 × 10⁻⁶ C m is aligned with a uniform field of 2.0 × 10⁵ N/C. An external agent turns it slowly to 90° from the field. What work does it do? (Answer: pE = 5.0 × 10⁻⁶ × 2.0 × 10⁵ = 1.0 J; W = pE (cos 0° − cos 90°) = 1.0 × (1 − 0) = 1.0 J.)
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Can you answer these Electrical Potential Energy Dipole Field MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In a uniform electric field E, the potential energy of a dipole of moment p at angle θ to the field, with the zero of energy taken at θ = 90°, is:
Show answer and why every option is right or wrong
Answer: C. Based on NCERT Class 12 Physics, Chapter 2, pages 60–61: the work stored as potential energy, with θ₀ = 90°, gives U(θ) = −pE cos θ.
Why A is wrong: A is wrong because pE sin θ has the form of the torque magnitude τ = pE sin θ, not of the energy.
Why B is wrong: B is wrong because it has the right trigonometric function but the wrong sign; with U = 0 at 90° the aligned dipole lies below zero.
Why D is wrong: D is wrong because it is a sine form, which belongs to torque; the energy involves the cosine of the angle.
MCQ 2Easy RecallPractice
NCERT takes θ₀ = 90° as the angle where the potential energy of a dipole is zero because, at that choice:
Show answer and why every option is right or wrong
Answer: A. Based on NCERT Class 12 Physics, Chapter 2, page 61: choosing θ₀ = 90° makes the work done against the field in bringing +q and −q equal and opposite, so they cancel.
Why B is wrong: B is wrong because the torque τ = pE sin θ is at its maximum at 90°, not zero; it vanishes at 0° and 180°.
Why C is wrong: C is wrong because NCERT states that in a uniform field the dipole feels no net force at any angle (page 60).
Why D is wrong: D is wrong because the stable position is θ = 0°, where U is least; 90° is neither an equilibrium nor a minimum.
MCQ 3Easy RecallPractice
A dipole is placed in a uniform electric field. Which statement is correct?
Show answer and why every option is right or wrong
Answer: D. Based on NCERT Class 12 Physics, Chapter 2, page 60: in a uniform field the net force on a dipole is zero, and the torque τ = p × E tends to rotate it unless p is parallel or antiparallel to E.
Why A is wrong: A is wrong because the two equal and opposite forces on +q and −q add to zero, and the torque is not zero for a general angle.
Why B is wrong: B is wrong because the net force vanishes in a uniform field; a net force needs a non-uniform field.
Why C is wrong: C is wrong because the torque is zero when p is parallel to E; it acts at every other angle.
MCQ 4Direct ApplicationPractice
A dipole of moment 2.0 × 10⁻⁶ C m lies along a uniform field of 5.0 × 10⁵ N/C. It is then turned through 60° from the field direction. The increase in its potential energy is:
Show answer and why every option is right or wrong
Answer: B. pE = 2.0 × 10⁻⁶ × 5.0 × 10⁵ = 1.0 J. U changes from −pE cos 0° = −1.0 J to −pE cos 60° = −0.50 J, so ΔU = +0.50 J (NCERT Class 12 Physics, Chapter 2, pages 60–61).
Why A is wrong: A is wrong because 1.5 J is pE (1 + cos 60°), which would be the change from the opposed position (180°) to 60°, not from the aligned position.
Why C is wrong: C is wrong because 0.87 J is pE sin 60°, the torque expression, used where the energy expression belongs.
Why D is wrong: D is wrong because 1.0 J is pE alone, the change from 0° to 90°; at 60° the cosine is 0.5 and not 0.
MCQ 5Direct ApplicationPractice
A dipole of moment 3.0 × 10⁻⁶ C m is held opposite to a uniform field of 4.0 × 10⁵ N/C (θ = 180°). With U = 0 at θ = 90°, its potential energy is:
Show answer and why every option is right or wrong
Answer: D. U = −pE cos 180° = +pE = 3.0 × 10⁻⁶ × 4.0 × 10⁵ = +1.2 J (NCERT Class 12 Physics, Chapter 2, pages 60–61).
Why A is wrong: A is wrong because −1.2 J is the energy at θ = 0°; at 180° cos θ = −1 and the sign flips.
Why B is wrong: B is wrong because zero energy belongs to θ = 90°, not to the opposed position.
Why C is wrong: C is wrong because +0.60 J takes half of pE, as if cos θ were 0.5; cos 180° has magnitude 1.
MCQ 6Direct ApplicationPractice
A dipole with pE = 2.0 J is turned very slowly by an external torque from θ₀ = 30° to θ₁ = 90°. The work done by the external torque is:
Show answer and why every option is right or wrong
Answer: A. W = pE (cos θ₀ − cos θ₁) = 2.0 × (cos 30° − cos 90°) = 2.0 × 0.866 = 1.7 J (NCERT Class 12 Physics, Chapter 2, page 60).
Why B is wrong: B is wrong because 1.0 J is pE (sin 90° − sin 30°); the work expression uses cosines, not sines.
Why C is wrong: C is wrong because 2.0 J is pE alone, which would need cos θ₀ = 1, that is θ₀ = 0°, with θ₁ = 90°.
Why D is wrong: D is wrong because the cosines are subtracted in the wrong order; the external torque moves the dipole away from the field, so its work is positive.
MCQ 7CalculationPractice
When a dipole is turned from alignment with a uniform field to 60° from it, its potential energy rises by 0.50 J. How much does its potential energy rise if it is turned from alignment to the opposite direction (180°)?
Show answer and why every option is right or wrong
Answer: C. ΔU(60°) = pE (1 − cos 60°) = pE/2 = 0.50 J, so pE = 1.0 J. Then ΔU(180°) = pE (1 − cos 180°) = 2pE = 2.0 J (NCERT Class 12 Physics, Chapter 2, pages 60–61).
Why A is wrong: A is wrong because 1.0 J is pE itself; the opposed position is 2pE above the aligned one, not pE.
Why B is wrong: B is wrong because 1.5 J assumes the rise is proportional to the angle (0.50 × 180/60); U depends on cos θ, so it is not linear in θ.
Why D is wrong: D is wrong because 0 J comes from using sin 180° = 0, the torque expression; the energy at 180° is +pE.
MCQ 8CalculationPractice
A dipole with pE = 4.0 J is held at 60° to a uniform field and then allowed to turn until it is aligned with the field. The change in its potential energy is:
Show answer and why every option is right or wrong
Answer: B. U at 60° = −pE cos 60° = −2.0 J; U at 0° = −pE = −4.0 J. U falls by 2.0 J (NCERT Class 12 Physics, Chapter 2, pages 60–61). NCERT's worked example on page 61 treats such a fall in energy as released, for instance as heat.
Why A is wrong: A is wrong because the dipole turns toward the field, to the lower-energy position, so U falls rather than rises.
Why C is wrong: C is wrong because 4.0 J is the final energy magnitude, not the change; the starting energy −2.0 J must be subtracted.
Why D is wrong: D is wrong because 6.0 J adds the 4.0 J final magnitude to the 2.0 J initial magnitude, instead of subtracting signed values.
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What to remember before solving Electrical Potential Energy Dipole Field questions
5 NCERT lines
This work is stored as the potential energy of the system. We can then associate potential energy U(q) with an inclination q of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take q0 = p / 2.
-- NCERT Class 12 Physics, Ch. 2, p. 60We can now understand why we took q0=p/2. In this case, the work done against the external field E in bringing +q and – q are equal and opposite and cancel out
-- NCERT Class 12 Physics, Ch. 2, p. 61Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J Change in potential energy = –3 J – (–6J) = 3 J So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.
-- NCERT Class 12 Physics, Ch. 2, p. 61Formula
Torque on dipole
τ = p × E; magnitude τ = p E sin θ where θ is angle between p and E. PE = -p·E = -p E cos θ.
-- NCERT Class 12 Physics, Ch. 1, p. 27Electrical Potential Energy Dipole Field: NEET previous year questions (PYQs) with answers
25 questions in Electrostatics
No question in our NEET 2020–2025 set targets this topic directly.
More in Electrostatics: 3 exam traps and mistakes · 11 formulas · 6 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Physics Chapter 2, p.60
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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