Electric Dipole and Potential Energy – Electrical potential energy of a system of two point: NEET MCQs with Answers

10 free questions on Electric Dipole and Potential Energy – Electrical potential energy of a system of two point for NEET (UG). Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Electrical potential energy of a system of two point charges and of electric dipole in an electrostatic field.

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Question 1

Two point charges are placed in an external electrostatic field: +3 μC at a point A where the external potential is 20 kV, and −2 μC at a point B where the external potential is 50 kV. A and B are 30 cm apart in air. Taking 1/4πε₀ = 9 × 10⁹ N m² C⁻², what is the total potential energy of the system?
  1. +0.14 J
  2. −0.22 J
  3. −0.18 J
  4. −0.04 J
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Answer: (B) −0.22 J
Explanation: The potential energy of the system is U = q₁V(A) + q₂V(B) + q₁q₂/(4πε₀r), which includes the mutual term. Here q₁V(A) = 3 × 10⁻⁶ × 2 × 10⁴ = +0.06 J and q₂V(B) = −2 × 10⁻⁶ × 5 × 10⁴ = −0.10 J. The mutual term is 9 × 10⁹ × (−6 × 10⁻¹²) / 0.3 = −0.18 J, so U = −0.22 J.

Question 2

An electric dipole consists of charges +5 nC and −5 nC separated by 4 mm. It is held in a uniform electric field of 3 × 10⁵ N/C in its orientation of unstable equilibrium. Taking the energy as zero when the dipole is perpendicular to the field, what are its potential energy and the torque on it?
  1. +12 × 10⁻⁶ J and zero torque
  2. +6 × 10⁻⁶ J and zero torque
  3. Zero energy and a torque of 6 × 10⁻⁶ N m
  4. −6 × 10⁻⁶ J and zero torque
Show answer
Answer: (B) +6 × 10⁻⁶ J and zero torque
Explanation: The dipole moment is p = (5 × 10⁻⁹)(4 × 10⁻³) = 2 × 10⁻¹¹ C m, so pE = 6 × 10⁻⁶ J. Unstable equilibrium occurs when p is opposite to E, that is θ = 180°. Then U = −pE cos 180° = +pE = +6 × 10⁻⁶ J, and the torque pE sin 180° is zero.

Question 3

Charges of +2 μC, −1 μC and +3 μC are initially very far apart. How much work must an external agent do to place them at the three corners of an equilateral triangle of side 30 cm, the charges ending at rest? (Take 1/(4πε₀) = 9 × 10⁹ N m² C⁻².)
  1. +30 mJ
  2. −30 mJ
  3. +330 mJ
  4. Zero
Show answer
Answer: (A) +30 mJ
Explanation: The work equals the potential energy of the final system, which is the sum over the three pairs: U = (1/4πε₀r)(q₁q₂ + q₂q₃ + q₃q₁). The bracket is (−2 − 3 + 6) × 10⁻¹² C² = +1 × 10⁻¹² C². So U = (9 × 10⁹/0.3) × 10⁻¹² J = 0.03 J = +30 mJ.

Question 4

A uniform external electric field of 1000 V/m acts along the +x direction, and the potential of this external field is taken as zero at x = 0. A charge of +2 μC is placed at x = 0 and a charge of −1 μC at x = 3 m. Taking 1/(4πε₀) = 9 × 10⁹ N m² C⁻², what is the total electrostatic potential energy of the pair, including their mutual interaction, in this field?
  1. +3 mJ
  2. −9 mJ
  3. −3 mJ
  4. −6 mJ
Show answer
Answer: (C) −3 mJ
Explanation: The total energy is q₁V(x₁) + q₂V(x₂) + q₁q₂/(4πε₀r). The external potential is V = −1000x, so V(0) = 0 and V(3 m) = −3000 V, giving 0 and (−1 μC)(−3000 V) = +3 mJ. The mutual term is 9 × 10⁹ × (2 × 10⁻⁶)(−1 × 10⁻⁶)/3 = −6 mJ. The total is −3 mJ.

Question 5

Two point charges of +2 μC each are fixed at points A and B, 12 m apart, and C is the midpoint of AB. A small particle of charge −2 μC is projected from C along the perpendicular bisector of AB. What is the least kinetic energy at C that lets it reach a point D on the bisector, 8 m from C? Neglect gravity. (1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. 19.2 mJ
  2. 7.2 mJ
  3. 4.8 mJ
  4. 12 mJ
Show answer
Answer: (C) 4.8 mJ
Explanation: The particle is 6 m from each fixed charge at C and √(6² + 8²) = 10 m from each at D. Its potential energy is U = −2kq²/r, with 2kq² = 0.072 J m, so U = −0.012 J at C and −0.0072 J at D. The rise in potential energy is 0.0048 J, which must be supplied by the kinetic energy at C.

Question 6

A dipole of moment p, free to turn about its centre without friction, is released from rest at an angle of 60° to a uniform electric field E. What is its maximum rotational kinetic energy in the motion that follows, and in which orientation does it occur?
  1. pE/2, when the dipole is perpendicular to the field
  2. pE, when the dipole is perpendicular to the field
  3. pE/2, when the dipole is aligned along the field
  4. (√3/2)pE, when the dipole is aligned along the field
Show answer
Answer: (C) pE/2, when the dipole is aligned along the field
Explanation: Potential energy is U = −pE cos θ, which is lowest at θ = 0, so kinetic energy peaks when the dipole lines up with the field. Energy conservation gives KE = U(60°) − U(0°) = −pE/2 + pE = pE/2. The dipole then swings on to −60° and oscillates about the aligned position.

Question 7

A dipole of moment 4 × 10⁻⁹ C m is in a uniform electric field of 5 × 10⁴ N/C. An external agent slowly turns it from an orientation at 60° to the field to one at 120° to the field. The torque has the same magnitude at both orientations. How much work does the external agent do?
  1. 3.5 × 10⁻⁴ J
  2. 1 × 10⁻⁴ J
  3. Zero
  4. 2 × 10⁻⁴ J
Show answer
Answer: (D) 2 × 10⁻⁴ J
Explanation: Work by the external agent equals the rise in potential energy, W = pE(cos θ₁ − cos θ₂). Here that is pE(cos 60° − cos 120°) = pE(½ + ½) = pE = 4 × 10⁻⁹ × 5 × 10⁴ = 2 × 10⁻⁴ J. Equal torque at the two ends does not make the work zero, because the energy depends on the angle through −pE cos θ.
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Question 8

An electric dipole in a uniform electric field has a potential energy of −3.0 × 10⁻⁶ J and experiences a torque of magnitude 4.0 × 10⁻⁶ N m in its present orientation. How much work must an external agent do to turn it slowly from this orientation to the orientation of maximum potential energy?
  1. 8.0 × 10⁻⁶ J
  2. 1.0 × 10⁻⁵ J
  3. 2.0 × 10⁻⁶ J
  4. 5.0 × 10⁻⁶ J
Show answer
Answer: (A) 8.0 × 10⁻⁶ J
Explanation: From U = −pE cosθ and τ = pE sinθ, we get pE cosθ = 3.0 × 10⁻⁶ J and pE sinθ = 4.0 × 10⁻⁶ J, so pE = 5.0 × 10⁻⁶ J. The maximum potential energy is +pE = +5.0 × 10⁻⁶ J. The work done slowly equals the rise in potential energy, 5.0 × 10⁻⁶ − (−3.0 × 10⁻⁶) = 8.0 × 10⁻⁶ J.

Question 9

Charges +Q, +Q and q are placed at the three vertices of an equilateral triangle. For the total electrostatic potential energy of the system to be zero, the value of q must be
  1. −Q/2
  2. −Q/4
  3. −2Q
  4. −Q
Show answer
Answer: (A) −Q/2
Explanation: All three pairs are the same distance apart, so the total energy is proportional to the sum of the pair products. This sum is Q² + Qq + Qq = Q² + 2Qq. Setting it to zero gives q = −Q/2.

Question 10

An electric dipole is in a uniform electric field. The work done by an external agent in slowly turning it from θ = 0° to θ = 60°, where θ is the angle between the dipole moment and the field, is W. The work needed to turn it further from θ = 60° to θ = 180° is
  1. W
  2. 4W
  3. 2W
  4. 3W
Show answer
Answer: (D) 3W
Explanation: The work to turn a dipole from θ₁ to θ₂ is pE(cos θ₁ − cos θ₂). From 0° to 60° this is pE/2 = W, so pE = 2W. From 60° to 180° it is pE(1/2 + 1) = 1.5pE = 3W.

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