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Electric charge, Coulomb’s law, multiple charges and superposition: NEET MCQs with answers

90 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. Consider the following two statements regarding electrostatic forces: Statement I: The presence of a third charge alters the electrostatic force exerted by one point charge on another. Statement II: The net force acting on any charge in a system of multiple charges is the vector sum of the forces exerted by all individual charges. Choose the correct option:

  1. Statement I is false but Statement II is true
  2. Both Statement I and Statement II are true
  3. Both Statement I and Statement II are false
  4. Statement I is true but Statement II is false
Show answer and explanation

Answer: (A) Statement I is false but Statement II is true

According to the principle of superposition, the interaction between any two charges is independent of the presence of other charges, so Statement I is false. The total force on any charge is indeed the vector sum of individual forces, so Statement II is true.

medium

2. Two point charges of 8 μC and 2 μC are placed 4 m apart in free space. Taking k = 9×10⁹ N m² C⁻², the magnitude of the electrostatic force between them is:

  1. 0.018 N
  2. 0.0045 N
  3. 0.009 N
  4. 0.036 N
Show answer and explanation

Answer: (C) 0.009 N

F = kq₁q₂/r² = 9×10⁹×(8×10⁻⁶)×(2×10⁻⁶)/4² = 0.009 N.

Quick tip: Convert μC to C and use the inverse-square relation.

hard

3. Two identical simple pendulums of string length l support equal masses m carrying equal charge q. In equilibrium in vacuum, the strings make an angle θ with the vertical, which is very small. The equilibrium separation x between the two charges is proportional to:

  1. q^(3/2)
  2. q^(1/2)
  3. q^(2/3)
  4. q²
Show answer and explanation

Answer: (C) q^(2/3)

For small angle θ, tan θ ≈ sin θ = (x/2) / l. Equilibrium condition gives tan θ = F_e / (mg) = (k q² / x²) / (mg). Thus, x / (2l) = k q² / (mg x²), which leads to x³ = 2k l q² / (mg). Therefore, x ∝ q^(2/3).

medium

4. Five identical point charges, each of charge +Q, are placed at five of the six vertices of a regular hexagon of side length L. What is the magnitude of the net electrostatic force exerted by these charges on a test charge +q placed at the center of the hexagon?

  1. √5 Qq / (4πε₀ L²)
  2. 5Qq / (4πε₀ L²)
  3. Qq / (4πε₀ L²)
  4. zero
Show answer and explanation

Answer: (C) Qq / (4πε₀ L²)

If a sixth charge +Q were present at the empty vertex, the net electrostatic force at the center would be zero due to symmetry. Therefore, the net force due to the five charges is equal in magnitude and opposite in direction to the force that would be exerted by a single charge +Q placed at the vacant vertex: F = Qq / (4πε₀ L²).

easy

5. Two point charges placed at a distance r in air experience a mutual force F. If they are placed at a distance r' in a medium of dielectric constant K = 16, they experience the same force F. The distance r' is equal to:

  1. 4r
  2. r/16
  3. r/4
  4. 16r
Show answer and explanation

Answer: (C) r/4

Coulomb's force in air is F = q₁q₂ / (4πε₀ r²). In the medium, the force is F' = q₁q₂ / (4πε₀ K r'²). For F' = F, we must have K r'² = r², so r' = r / √K = r / √16 = r/4.

hard

6. Two identical charged spheres suspended from a common point by strings of equal length hang at an angle 2θ with each other in air. When suspended in a liquid of density 0.8 g cm⁻³, the angle between the strings remains unchanged. If the density of the material of the spheres is 1.6 g cm⁻³, the dielectric constant of the liquid is:

  1. 4
  2. 2
  3. 1.5
  4. 2.5
Show answer and explanation

Answer: (B) 2

In air, tan θ = F / (mg). In liquid, tan θ = F' / (mg(1 - d_l/d_s)) where F' = F / K. Since θ remains the same, F / (mg) = F / [K mg (1 - d_l / d_s)], which simplifies to K = 1 / (1 - d_l/d_s). Substituting values: K = 1 / (1 - 0.8/1.6) = 1 / (1 - 0.5) = 2.

medium

7. Two point charges of 8 μC and 2 μC are 40 cm apart in vacuum. Taking 1/(4πε₀) = 9 × 10⁹ N m² C⁻², the force between them is:

  1. 9 N
  2. 0.9 N
  3. 0.36 N
  4. 0.09 N
Show answer and explanation

Answer: (B) 0.9 N

F = kq₁q₂/r² = 9 × 10⁹ × 8 × 10⁻⁶ × 2 × 10⁻⁶/(0.4)² = 0.9 N.

medium

8. Two point charges of 1 μC and 4 μC are 20 cm apart in vacuum. Taking 1/(4πε₀) = 9 × 10⁹ N m² C⁻², the force between them is:

  1. 0.09 N
  2. 9 N
  3. 0.18 N
  4. 0.9 N
Show answer and explanation

Answer: (D) 0.9 N

F = kq₁q₂/r² = 9 × 10⁹ × 1 × 10⁻⁶ × 4 × 10⁻⁶/(0.2)² = 0.9 N.

medium

9. Two point charges of 3 μC and 7 μC are placed 2 m apart in vacuum. Taking k = 9×10⁹ N m² C⁻², the magnitude of the electrostatic force between them is:

  1. 0.0473 N
  2. 0.0237 N
  3. 0.0946 N
  4. 0.0945 N
Show answer and explanation

Answer: (A) 0.0473 N

F = kq₁q₂/r² = 9×10⁹×(3×10⁻⁶)×(7×10⁻⁶)/2² = 0.0473 N.

Quick tip: Convert μC to C and use the inverse-square relation.

medium

10. Two point charges carrying 3 μC and 1 μC are separated by 3 m in vacuum. With k = 9×10⁹ N m² C⁻², the electrostatic force between them is:

  1. 0.003 N
  2. 0.009 N
  3. 0.0015 N
  4. 0.006 N
Show answer and explanation

Answer: (A) 0.003 N

F = kq₁q₂/r² = 9×10⁹×(3×10⁻⁶)×(1×10⁻⁶)/3² = 0.003 N.

Quick tip: Convert μC to C and use the inverse-square relation.

medium

11. Two point charges of 8 μC and 3 μC are 20 cm apart in vacuum. Taking 1/(4πε₀) = 9 × 10⁹ N m² C⁻², the force between them is:

  1. 1.08 N
  2. 0.54 N
  3. 54 N
  4. 5.4 N
Show answer and explanation

Answer: (D) 5.4 N

F = kq₁q₂/r² = 9 × 10⁹ × 8 × 10⁻⁶ × 3 × 10⁻⁶/(0.2)² = 5.4 N.

medium

12. Two point-sized charged bodies carrying 3 μC and 7 μC are separated by 2 m in free space. With k = 9×10⁹ N m² C⁻², the electrostatic force between them is:

  1. 0.0473 N
  2. 0.0237 N
  3. 0.0946 N
  4. 0.0945 N
Show answer and explanation

Answer: (A) 0.0473 N

F = kq₁q₂/r² = 9×10⁹×(3×10⁻⁶)×(7×10⁻⁶)/2² = 0.0473 N.

Quick tip: Convert μC to C and use the inverse-square relation.

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