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Electric field, field lines and continuous charge distributions: NEET MCQs with answers

75 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. Which one of the following statements about electrostatic field lines is correct?

  1. They can form closed loops in the presence of static point charges.
  2. Tangent to an electric field line at any point gives the direction of the electric force on a negative charge at that point.
  3. The relative closeness of field lines in a region indicates the magnitude of electric field intensity in that region.
  4. Electric field lines can intersect each other if the charges are of opposite sign.
Show answer and explanation

Answer: (C) The relative closeness of field lines in a region indicates the magnitude of electric field intensity in that region.

Electric field lines never form closed loops in electrostatics because the electrostatic field is conservative. Tangent at any point gives the direction of the electric field, which is the force on a positive test charge (force on a negative charge is opposite). The crowding or density of field lines represents the magnitude of the field. Field lines never cross each other.

medium

2. A charged water droplet of mass 9.6 × 10⁻¹⁵ kg is held stationary between two horizontal plates separated by 2.0 cm in an electric field. If the droplet carries an excess of 3 electrons (e = 1.6 × 10⁻¹⁹ C, g = 10 m s⁻²), the magnitude of the electric field required to balance its weight is:

  1. 4.8 × 10⁴ N C⁻¹
  2. 1.5 × 10⁵ N C⁻¹
  3. 2.0 × 10⁵ N C⁻¹
  4. 3.2 × 10⁴ N C⁻¹
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Answer: (C) 2.0 × 10⁵ N C⁻¹

For the droplet to remain stationary, the upward electrostatic force balances gravity: qE = mg. Here q = 3e = 3 × 1.6 × 10⁻¹⁹ C = 4.8 × 10⁻¹⁹ C, and m = 9.6 × 10⁻¹⁵ kg. Thus, E = mg / q = (9.6 × 10⁻¹⁵ × 10) / (4.8 × 10⁻¹⁹) = 2.0 × 10⁵ N C⁻¹.

hard

3. An electric field line emerges from a positive charge q₁ at an angle α to the straight line joining q₁ and a negative charge -q₂ (where q₁ > |q₂|). The field line enters the charge -q₂ at an angle β. Which of the following relations is correct?

  1. q₁ sin²(α/2) = |q₂| sin²(β/2)
  2. q₁ sin(α/2) = |q₂| sin(β/2)
  3. q₁ cos(α/2) = |q₂| cos(β/2)
  4. q₁ tan(α) = |q₂| tan(β)
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Answer: (A) q₁ sin²(α/2) = |q₂| sin²(β/2)

The number of field lines in a cone of semi-vertical angle θ around a point charge is proportional to the solid angle Ω = 2π(1 - cos θ) = 4π sin²(θ/2). By conservation of field lines passing from q₁ into -q₂, q₁ × 2π(1 - cos α) = |q₂| × 2π(1 - cos β), which simplifies directly to q₁ sin²(α/2) = |q₂| sin²(β/2).

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4. Given below are two statements: Statement I: A small charged particle released from rest in an electric field will always move strictly along the electric field line passing through that point. Statement II: The electric field line gives the direction of acceleration of a positively charged particle at any instant. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
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Answer: (C) Statement I is incorrect but Statement II is correct.

Statement I is incorrect because if the field lines are curved, the particle's inertia carries it off the field line as soon as it acquires velocity. Statement II is correct because F = qE = ma, so the acceleration of a positive charge is always in the direction of the field vector E, which is tangent to the field line.

easy

5. The magnitude of the electric field 2 m from a point charge of 7 μC in vacuum (1/(4πε₀) = 9 × 10⁹ N m² C⁻²) is:

  1. 31500 N/C
  2. 7875 N/C
  3. 1.57 × 10¹⁰ N/C
  4. 15750 N/C
Show answer and explanation

Answer: (D) 15750 N/C

E = kq/r² = 9 × 10⁹ × 7 × 10⁻⁶/2² = 15750 N/C.

hard

6. A uniformly charged hemispherical shell of radius R has a total charge Q. The electric field at the center of curvature of the hemisphere has a magnitude of:

  1. Q / (8πε₀ R²)
  2. Q / (4πε₀ R²)
  3. Q / (16πε₀ R²)
  4. Q / (2πε₀ R²)
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Answer: (A) Q / (8πε₀ R²)

The surface charge density is σ = Q / (2πR²). Integrating rings of charge over the hemisphere gives the field at the center of curvature as E = σ / (4ε₀). Substituting σ = Q / (2πR²) yields E = Q / (8πε₀ R²).

easy

7. The electric field at a distance x on the axis of a uniformly charged thin ring of radius R carrying a total charge Q is maximum at a distance x equal to:

  1. R / 2
  2. R
  3. R / √2
  4. √2 R
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Answer: (C) R / √2

The axial electric field of a ring is given by E = (1 / 4πε₀) · Qx / (R² + x²)^(3/2). Setting dE/dx = 0 yields (R² + x²)^(3/2) - x · (3/2)(R² + x²)^(1/2)(2x) = 0, which gives R² - 2x² = 0, leading to x = R / √2.

easy

8. Two point charges +q and +9q are placed at a distance L apart. At what distance from the charge +q on the line joining them is the net electric field zero?

  1. L / 4
  2. L / 3
  3. L / 2
  4. 3L / 4
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Answer: (A) L / 4

Let the point be at distance x from +q. For the net field to be zero, the individual fields must balance: kq / x² = k(9q) / (L - x)². Taking square roots on both sides gives 1/x = 3/(L - x), which leads to L - x = 3x, or x = L/4.

easy

9. The magnitude of the electric field 1 m from a point charge of 10 μC in vacuum (1/(4πε₀) = 9 × 10⁹ N m² C⁻²) is:

  1. 1.8 × 10⁵ N/C
  2. 90000 N/C
  3. 45000 N/C
  4. 9 × 10¹⁰ N/C
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Answer: (B) 90000 N/C

E = kq/r² = 9 × 10⁹ × 10 × 10⁻⁶/1² = 90000 N/C.

easy

10. Two point charges +16 μC and -9 μC are placed 7 cm apart in air. At what point on the line passing through both charges is the net electric field zero?

  1. 4 cm from the +16 μC charge, between the charges
  2. 3 cm from the -9 μC charge, between the charges
  3. 21 cm from the -9 μC charge, outside the segment joining the charges
  4. 28 cm from the +16 μC charge, between the charges
Show answer and explanation

Answer: (C) 21 cm from the -9 μC charge, outside the segment joining the charges

For charges of opposite signs, the neutral point lies outside the charges on the line joining them, closer to the smaller magnitude charge (-9 μC). Let the distance from -9 μC be d. Then 16/(7 + d)² = 9/d², which gives 4/(7 + d) = 3/d, leading to 4d = 21 + 3d, so d = 21 cm from the -9 μC charge.

easy

11. The magnitude of the electric field 0.5 m from a point charge of 10 μC in vacuum (1/(4πε₀) = 9 × 10⁹ N m² C⁻²) is:

  1. 3.6 × 10⁵ N/C
  2. 7.2 × 10⁵ N/C
  3. 1.8 × 10⁵ N/C
  4. 3.6 × 10¹¹ N/C
Show answer and explanation

Answer: (A) 3.6 × 10⁵ N/C

E = kq/r² = 9 × 10⁹ × 10 × 10⁻⁶/0.5² = 3.6 × 10⁵ N/C.

easy

12. The magnitude of the electric field 2 m from a point charge of 6 μC in vacuum (1/(4πε₀) = 9 × 10⁹ N m² C⁻²) is:

  1. 27000 N/C
  2. 1.35 × 10¹⁰ N/C
  3. 6750 N/C
  4. 13500 N/C
Show answer and explanation

Answer: (D) 13500 N/C

E = kq/r² = 9 × 10⁹ × 6 × 10⁻⁶/2² = 13500 N/C.

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