Electric Dipole: NEET MCQs with Answers

15 free questions on Electric Dipole for NEET (UG). Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Electric dipole, Electric field due to a dipole. Torque on a dipole in a uniform electric field.

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Question 1

CO₂ and H₂O are both triatomic molecules, but only H₂O has a permanent electric dipole moment. What is the reason?
  1. An H₂O molecule carries a net positive charge, whereas a CO₂ molecule is neutral
  2. CO₂ is a gas at room temperature, whereas water is a liquid
  3. Carbon and oxygen have the same electronegativity, so charge is not displaced in the bonds of CO₂
  4. CO₂ is linear and symmetric, so its centres of positive and negative charge coincide, whereas in the bent H₂O molecule they do not
Show answer
Answer: (D) CO₂ is linear and symmetric, so its centres of positive and negative charge coincide, whereas in the bent H₂O molecule they do not
Explanation: A molecule has a permanent dipole moment only if its centres of positive and negative charge are separated. In linear CO₂ the two bond dipoles are equal and opposite and cancel. In bent H₂O the bond dipoles do not cancel, so the two centres stay apart even without a field.

Question 2

An electric dipole of moment 4 × 10⁻⁹ C m is held near a large plane non-conducting sheet with uniform surface charge density 1.77 × 10⁻⁶ C/m². The dipole axis makes an angle of 30° with the normal to the sheet. The torque on the dipole is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²):
  1. 1.0 × 10⁻⁴ N m
  2. 3.5 × 10⁻⁴ N m
  3. 2.0 × 10⁻⁴ N m
  4. 4.0 × 10⁻⁴ N m
Show answer
Answer: (C) 2.0 × 10⁻⁴ N m
Explanation: The field of a large sheet is uniform, E = σ/2ε₀ = 1.77 × 10⁻⁶ / (1.77 × 10⁻¹¹) = 1.0 × 10⁵ N/C, along the normal and independent of distance. The torque is τ = pE sin θ = 4 × 10⁻⁹ × 1.0 × 10⁵ × 0.5 = 2.0 × 10⁻⁴ N m.

Question 3

Four charges of magnitude q are placed at the corners of a square of side a. In arrangement X, two adjacent corners carry +q and the other two carry −q. In arrangement Y, the signs alternate around the square. The magnitudes of the net electric dipole moments of X and Y are respectively
  1. 2qa and 0
  2. 2√2 qa and 0
  3. 2qa and 2√2 qa
  4. 0 and 2qa
Show answer
Answer: (A) 2qa and 0
Explanation: In X, the charges form two parallel dipoles of moment qa each, which add to 2qa. In Y, take corners (0, 0) and (a, a) as +q, with (a, 0) and (0, a) as −q. The vector sum of charge times position is q[(a, a) − (a, 0) − (0, a)] = 0, so the net dipole moment vanishes.

Question 4

A short electric dipole of moment p = (2î + ĵ − k̂) × 10⁻²⁹ C m is at the origin. Which vector gives the exact direction, including sense, of the electric field at the far point with position vector r = 10î − 10ĵ + 10k̂ m?
  1. 2î + ĵ − k̂
  2. −î + ĵ − k̂
  3. î − ĵ + k̂
  4. −2î − ĵ + k̂
Show answer
Answer: (D) −2î − ĵ + k̂
Explanation: Here p · r = 20 − 10 − 10 = 0, so the point lies on the equatorial plane of the dipole. At an equatorial point the field is antiparallel to the dipole moment. Its direction is therefore −p, that is, along −2î − ĵ + k̂.

Question 5

Charges +2q, +q and −3q are placed at (a, 0), (0, a) and (0, 0) respectively. What is the magnitude of the electric dipole moment of this system, and how does it depend on the choice of origin?
  1. √5 qa, and it does not depend on the origin
  2. √5 qa, and it changes when the origin is shifted
  3. 3qa, and it does not depend on the origin
  4. 3qa, and it changes when the origin is shifted
Show answer
Answer: (A) √5 qa, and it does not depend on the origin
Explanation: The dipole moment of a system of charges is p = Σqᵢrᵢ, which here is 2qa î + qa ĵ, of magnitude √5 qa. Shifting the origin by a vector d changes p by −(Σqᵢ)d. Since the net charge is zero, p is unchanged.

Question 6

The electric field in a region is E⃗ = Cx î, where C = 2 × 10⁶ N C⁻¹ m⁻¹. A short dipole of moment 3 × 10⁻⁹ C m lies on the positive x-axis with its dipole moment pointing along −x. What net force acts on it?
  1. 6 × 10⁻³ N along −x
  2. 3 × 10⁻³ N along −x
  3. Zero, because its net charge is zero
  4. 6 × 10⁻³ N along +x
Show answer
Answer: (A) 6 × 10⁻³ N along −x
Explanation: With p⃗ along −x, the +q charge lies at x₀ − a and the −q charge at x₀ + a, so the two charges feel fields of different strength. The net force along x is qC(x₀ − a) − qC(x₀ + a) = −2qaC = −pC. Its magnitude is 3 × 10⁻⁹ × 2 × 10⁶ = 6 × 10⁻³ N along −x, so a dipole opposed to the field is pushed towards the weaker field region.

Question 7

A small electric dipole of moment p⃗ = (2î − ĵ + 2k̂) × 10⁻²⁹ C m is at the origin. Which vector gives the exact direction, including sense, of its electric field at the point r⃗ = (î + 2ĵ) m?
  1. −4î + 2ĵ + 5k̂
  2. 2î − ĵ + 2k̂
  3. −2î + ĵ − 2k̂
  4. î + 2ĵ
Show answer
Answer: (C) −2î + ĵ − 2k̂
Explanation: Here r⃗·p⃗ = 2 − 2 + 0 = 0, so the point lies on the equatorial plane of the dipole. On this plane the field is E⃗ = −p⃗/(4πε₀r³), which is antiparallel to the dipole moment. The direction is therefore −2î + ĵ − 2k̂.
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Question 8

An ammonia molecule has 10 protons and 10 electrons, and its dipole moment in the vapour state is 4.8 × 10⁻³⁰ C m. If all the positive charge is treated as acting at one centre and all the negative charge at another, what is the distance between these two centres? (e = 1.6 × 10⁻¹⁹ C)
  1. 3 × 10⁻¹⁰ m
  2. 3 × 10⁻¹² m
  3. 3 × 10⁻¹¹ m
  4. 6 × 10⁻¹² m
Show answer
Answer: (B) 3 × 10⁻¹² m
Explanation: The charge at each centre is q = 10e = 1.6 × 10⁻¹⁸ C. Since p = qd, the separation is d = p/q = 4.8 × 10⁻³⁰/1.6 × 10⁻¹⁸ = 3 × 10⁻¹² m. This is far smaller than the size of the molecule, because the two charge centres nearly coincide.

Question 9

For a short electric dipole, compare two points at the same large distance r from its centre: point A, where the line from the centre to A makes 60° with the dipole axis, and point B, on the equatorial line. What is the ratio of the field magnitude at A to that at B?
  1. √7/2
  2. √3/2
  3. 2
  4. 7/4
Show answer
Answer: (A) √7/2
Explanation: At angle θ from the axis the field has a radial part 2kp cos θ/r³ and a transverse part kp sin θ/r³, so its magnitude is (kp/r³)√(1 + 3cos²θ). At 60° this gives (kp/r³)√(1 + 3/4) = (√7/2)(kp/r³). At the equatorial point θ = 90°, so the field is kp/r³, and the ratio is √7/2.

Question 10

Charges +5 nC and −5 nC are held 6 cm apart. Point P lies on the perpendicular bisector of the line joining them, 4 cm from its midpoint. What is the electric field at P? (1/4πε₀ = 9 × 10⁹ N m² C⁻²)
  1. 4.2 × 10⁴ N/C, directed opposite to the dipole moment
  2. 2.16 × 10⁴ N/C, directed opposite to the dipole moment
  3. 2.16 × 10⁴ N/C, directed along the dipole moment
  4. 2.88 × 10⁴ N/C, directed opposite to the dipole moment
Show answer
Answer: (B) 2.16 × 10⁴ N/C, directed opposite to the dipole moment
Explanation: P is 5 cm from each charge, since √(3² + 4²) = 5, so each charge gives a field of 9 × 10⁹ × 5 × 10⁻⁹ / (0.05)² = 1.8 × 10⁴ N/C. The components along the bisector cancel, and the components parallel to the line joining the charges add to 2 × 1.8 × 10⁴ × (3/5) = 2.16 × 10⁴ N/C. This net field points from the positive towards the negative charge, opposite to the dipole moment. The short-dipole formula cannot be used here because the separation is not small compared with the distance.

Question 11

An electric field in a region points along +x, and its magnitude increases uniformly along +x at 5 × 10⁴ N C⁻¹ m⁻¹. A small electric dipole of moment 2 × 10⁻⁶ C m lies along the x-axis with its dipole moment pointing along −x. What force and torque act on it?
  1. Force 0.05 N along −x; torque zero
  2. Force zero; torque 0.1 N m
  3. Force 0.1 N along −x; torque zero
  4. Force 0.1 N along +x; torque zero
Show answer
Answer: (C) Force 0.1 N along −x; torque zero
Explanation: The force magnitude is p(dE/dx) = 2 × 10⁻⁶ × 5 × 10⁴ = 0.1 N. With the moment along −x, the negative charge sits at larger x, where the field is stronger, so its force along −x wins over the force along +x on the positive charge. The dipole axis lies along the field, so sin θ = 0 and the torque is zero.

Question 12

In a region the electric field is E = (5 × 10⁵ x) î N/C, with x in metres. A short dipole of moment 4 × 10⁻⁹ C m is held at a point with x > 0, with its moment pointing along the −x direction. The net force on the dipole is:
  1. Zero, since the total charge of a dipole is zero
  2. 2 × 10⁻³ N along −x
  3. 2 × 10⁻³ N along +x
  4. 4 × 10⁻³ N along −x
Show answer
Answer: (B) 2 × 10⁻³ N along −x
Explanation: The field grows with x, so the two charges feel unequal forces. The negative charge sits at the larger x, in the stronger field, and is pushed along −x more strongly than the positive charge is pushed along +x. The net force has magnitude p(dE/dx) = 4 × 10⁻⁹ × 5 × 10⁵ = 2 × 10⁻³ N, directed along −x.

Question 13

A dipole consists of +2 μC at (3, 0, 2) m and −2 μC at (1, 2, 1) m. It is placed in a uniform electric field E = 10 k̂ N/C. The torque on the dipole, in N m, is:
  1. −4 × 10⁻⁵ (î + ĵ)
  2. −4 × 10⁻⁵ (î − ĵ)
  3. +4 × 10⁻⁵ (î − ĵ)
  4. +4 × 10⁻⁵ (î + ĵ)
Show answer
Answer: (A) −4 × 10⁻⁵ (î + ĵ)
Explanation: The displacement from the negative to the positive charge is 2î − 2ĵ + k̂ metres, so p = (4î − 4ĵ + 2k̂) × 10⁻⁶ C m. Then τ = p × E, and using î × k̂ = −ĵ and ĵ × k̂ = î gives (−40î − 40ĵ) × 10⁻⁶ N m. This equals −4 × 10⁻⁵ (î + ĵ) N m.

Question 14

A bent triatomic molecule is modelled as a charge −2q on the central atom and a charge +q on each of two outer atoms. Each outer atom is at distance d from the central atom, and the angle between the two bonds is 120°. What is the magnitude of the dipole moment of this model?
  1. 2qd
  2. √3 qd
  3. qd
  4. Zero
Show answer
Answer: (C) qd
Explanation: The system is neutral, so it can be treated as two dipoles of moment qd, each pointing from the central atom to an outer atom. These two moments are 120° apart. Their resultant is 2qd cos 60° = qd, directed along the bisector of the bond angle towards the positive charges.

Question 15

A charge of −2 μC is at point A (0, 1, 1) m and a charge of +2 μC is at point B (3, 1, 5) m. The pair is placed in a uniform electric field E = 5 ĵ V/cm. What is the magnitude of the torque on this dipole?
  1. 2.5 × 10⁻³ N m
  2. 5.0 × 10⁻³ N m
  3. 7.0 × 10⁻³ N m
  4. 5.0 × 10⁻⁵ N m
Show answer
Answer: (B) 5.0 × 10⁻³ N m
Explanation: The displacement from A to B is (3î + 4k̂) m, so p = 2 × 10⁻⁶ (3î + 4k̂) C m, of magnitude 10 × 10⁻⁶ C m. The field is 5 V/cm = 500 V/m along ĵ, which is perpendicular to p. Hence τ = pE = 10 × 10⁻⁶ × 500 = 5.0 × 10⁻³ N m.

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