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Electric dipole, dipole field and torque: NEET MCQs with answers

50 questions on this topic of Electrostatics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. An electric dipole of moment p is placed in a uniform external electric field E. If θ is the angle between the dipole moment vector p and the electric field vector E, which of the following orientations corresponds to the dipole being in unstable equilibrium?

  1. θ = 0°
  2. θ = 180°
  3. θ = 90°
  4. θ = 270°
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Answer: (B) θ = 180°

The torque on a dipole is τ = p × E = pE sin θ. At both θ = 0° and θ = 180°, the net torque is zero. The potential energy is U = -p · E = -pE cos θ. At θ = 180°, U = +pE, which is maximum potential energy, corresponding to unstable equilibrium.

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2. An electric dipole of length 4 cm, when placed with its axis at 30° to a uniform electric field of 10⁵ N C⁻¹, experiences a torque of 4√3 N m. The magnitude of each charge of the dipole is:

  1. 2.0 × 10⁻³ C
  2. 1.73 × 10⁻³ C
  3. 4.0 × 10⁻³ C
  4. 3.46 × 10⁻³ C
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Answer: (D) 3.46 × 10⁻³ C

Torque is τ = pE sin θ = (q × 2a)E sin θ. Given τ = 4√3 N m, 2a = 0.04 m, E = 10⁵ N C⁻¹, and θ = 30°. So, 4√3 = q × 0.04 × 10⁵ × 0.5 = 2000 q. Thus, q = 4√3 / 2000 = 2√3 × 10⁻³ C ≈ 3.46 × 10⁻³ C.

hard

3. An electric dipole has charges +q and -q separated by a fixed distance 2a. A point P lies on the perpendicular bisector of the dipole at a distance y from its centre. What is the EXACT expression for the magnitude of the electric field at point P?

  1. qy / [4πε₀ (y² + a²)³ᐟ²]
  2. q / [4πε₀ (y² + a²)]
  3. 2qa / [4πε₀ (y² - a²)³ᐟ²]
  4. 2qa / [4πε₀ (y² + a²)³ᐟ²]
Show answer and explanation

Answer: (D) 2qa / [4πε₀ (y² + a²)³ᐟ²]

Each charge is at a distance √(y² + a²) from point P. The magnitude of the field due to each charge is E₀ = q / [4πε₀ (y² + a²)]. The components perpendicular to the dipole axis cancel, while the components parallel to the dipole axis add up: E_net = 2 E₀ cos ϕ, where cos ϕ = a / √(y² + a²). Hence, E_net = 2q a / [4πε₀ (y² + a²)³ᐟ²].

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4. A molecule has an electric dipole moment of 2 × 10⁻²⁹ C m. It is initially aligned in a uniform electric field of magnitude 5 × 10⁴ V m⁻¹. The work required to completely reverse its orientation from stable to unstable equilibrium is:

  1. 1 × 10⁻²⁴ J
  2. 0 J
  3. 4 × 10⁻²⁴ J
  4. 2 × 10⁻²⁴ J
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Answer: (D) 2 × 10⁻²⁴ J

Stable equilibrium occurs at θ = 0° (U₁ = -pE) and unstable equilibrium occurs at θ = 180° (U₂ = +pE). The work required to reverse the direction is W = U₂ - U₁ = 2pE = 2 × (2 × 10⁻²⁹) × (5 × 10⁴) = 2 × 10⁻²⁴ J.

easy

5. An electric dipole of moment p is held in a uniform electric field E such that it makes an angle of 30° with the field. If the torque acting on it has a magnitude τ, what will be the magnitude of torque when its orientation is changed so that it makes an angle of 60° with the field?

  1. √3 τ
  2. τ / √3
  3. τ / 2
  4. 2τ
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Answer: (A) √3 τ

The magnitude of torque on a dipole in a uniform electric field is τ = pE sin θ. For θ = 30°, τ = pE sin 30° = pE / 2, giving pE = 2τ. For θ = 60°, τ' = pE sin 60° = (2τ)(√3 / 2) = √3 τ.

hard

6. A short electric dipole is placed at the origin with its dipole moment p directed along the +z-axis. At a point P having spherical polar coordinates (r, θ), where θ is the polar angle with the z-axis and r is very large, the electric field vector E makes an angle α with the position vector r such that tan α = (1/2) tan θ. For what value of θ is the electric field E perpendicular to the dipole moment vector p?

  1. θ = 0°
  2. θ = 90°
  3. θ = arccos(1/√3)
  4. θ = arcsin(1/√3)
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Answer: (C) θ = arccos(1/√3)

The total angle that E makes with the z-axis is (θ + α). For E to be perpendicular to the z-axis (the dipole direction), we must have θ + α = 90°, which means α = 90° - θ. Therefore, tan α = cot θ. Using the dipole field relation tan α = (1/2) tan θ, we get cot θ = (1/2) tan θ, which implies tan² θ = 2. Since sec² θ = 1 + tan² θ = 3, we have cos² θ = 1/3, or cos θ = 1/√3. Thus, θ = arccos(1/√3).

easy

7. A short electric dipole has a dipole moment of 4 × 10⁻⁹ C m. What is the magnitude of the electric potential at a point situated at a distance of 0.3 m from the centre of the dipole on a line making an angle of 60° with the dipole axis? (Take 1/(4πε₀) = 9 × 10⁹ N m² C⁻²)

  1. 100 V
  2. 200 V
  3. 400 V
  4. 50 V
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Answer: (B) 200 V

The potential due to a short dipole at distance r and angle θ is V = (1/(4πε₀)) × (p cos θ / r²). Substituting the given values: V = (9 × 10⁹) × (4 × 10⁻⁹ × cos 60°) / (0.3)² = (36 × 0.5) / 0.09 = 18 / 0.09 = 200 V.

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8. A small electric dipole is oriented along the +z-axis. The electric field is non-uniform and points in the +z-direction, increasing uniformly along the +z-direction at a rate of 10⁴ V m⁻² (i.e., dE/dz = +10⁴ N C⁻¹ m⁻¹). If the dipole moment is 2 × 10⁻⁷ C m in the +z-direction, what is the net electrostatic force experienced by the dipole?

  1. 2 × 10⁻³ N in the +z-direction
  2. 2 × 10⁻³ N in the -z-direction
  3. 4 × 10⁻³ N in the +z-direction
  4. zero
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Answer: (A) 2 × 10⁻³ N in the +z-direction

For a dipole aligned along the field gradient, the net force is given by F = p × (dE/dz). Here, p = 2 × 10⁻⁷ C m and dE/dz = 10⁴ N C⁻¹ m⁻¹. Hence, F = (2 × 10⁻⁷) × 10⁴ = 2 × 10⁻³ N. Since the field increases along +z and the positive charge is at a higher z-coordinate than the negative charge, the net force is along the +z-direction.

easy

9. Which of the following statements is INCORRECT regarding an electric dipole?

  1. The net electric charge of an electric dipole is zero.
  2. The electric field of a short dipole at equatorial points is perpendicular to the dipole axis.
  3. The total work done in rotating a dipole through 360° in a uniform electric field is zero.
  4. The electric field at every point in space due to a dipole is non-zero.
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Answer: (B) The electric field of a short dipole at equatorial points is perpendicular to the dipole axis.

The electric field of a short dipole at equatorial points is directed antiparallel to the dipole moment, which means it is parallel to the dipole axis, not perpendicular to it.

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10. A point charge +Q is fixed at a large distance r from the centre of a short dipole of moment p on its equatorial plane. If the dipole moment vector p is perpendicular to the line joining the dipole centre and the charge Q, what is the magnitude of the electrostatic force exerted by the dipole on the charge Q?

  1. Qp / (2πε₀ r⁴)
  2. 2Qp / (4πε₀ r³)
  3. Qp / (4πε₀ r³)
  4. Zero
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Answer: (C) Qp / (4πε₀ r³)

By Newton's third law, the force on the point charge due to the dipole is equal in magnitude to the force on the dipole due to the charge. At a distance r along the equatorial line, the electric field of the short dipole has magnitude E = p / (4πε₀ r³). The electrostatic force on the charge Q is F = Q E = Qp / (4πε₀ r³).

medium

11. The electric field due to a dipole at large distances falls off as:

  1. 1/r⁴
  2. 1/r³
  3. 1/r²
  4. 1/r
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Answer: (B) 1/r³

Because the dipole's total charge is zero, its far field falls faster than that of a point charge: as 1/r³ rather than 1/r².

NCERT Class 12

easy

12. The electric dipole moment of charges +q and -q separated by 2a is:

  1. Zero
  2. 2qa, directed from -q to +q
  3. 2qa, directed from +q to -q
  4. qa, directed from -q to +q
Show answer and explanation

Answer: (B) 2qa, directed from -q to +q

Dipole moment p = q × 2a. By convention its direction is from the negative charge to the positive charge.

NCERT Class 12

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