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Electric flux and Gauss’s law with standard applications: 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. A point charge q is placed at a distance a/2 directly above the centre of a horizontal square surface of side length a. The electric flux passing through this square surface is:

  1. q / (6ε₀)
  2. q / (4ε₀)
  3. q / ε₀
  4. q / (8ε₀)
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Answer: (A) q / (6ε₀)

By symmetry, the square surface of side a can be considered as one of the six identical faces of a fictitious cube of edge length a with the charge q positioned exactly at its centre. By Gauss's law, total flux through the cube is q/ε₀, hence the flux through one face is q/(6ε₀).

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2. An electric dipole consisting of charges +2 μC and -2 μC separated by a distance of 3 cm is placed inside a closed spherical surface of radius 10 cm. Additionally, a point charge of +5 μC is located inside the sphere, while a charge of -7 μC is located outside the sphere. The net electric flux emerging through the spherical surface is:

  1. -2.26 × 10⁵ N·m²/C
  2. 2.26 × 10⁵ N·m²/C
  3. 5.65 × 10⁵ N·m²/C
  4. 0
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Answer: (C) 5.65 × 10⁵ N·m²/C

According to Gauss's law, Φ = q_enclosed / ε₀. The total charge enclosed inside the surface is q_enclosed = (+2 μC) + (-2 μC) + (+5 μC) = +5 μC = 5 × 10⁻⁶ C (charges outside do not contribute). Thus, Φ = (5 × 10⁻⁶ C) / (8.854 × 10⁻¹² C²/N·m²) ≈ 5.65 × 10⁵ N·m²/C.

hard

3. A total charge Q is distributed uniformly throughout the volume of a non-conducting sphere of radius R. A spherical cavity of radius R/2 is scooped out such that its surface touches the outer surface of the sphere and passes through the centre of the sphere. The electric field at the centre of the cavity is:

  1. 3Q / (16πε₀R²)
  2. Q / (16πε₀R²)
  3. Q / (32πε₀R²)
  4. Q / (8πε₀R²)
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Answer: (D) Q / (8πε₀R²)

By superposition, E_cavity = E_complete - E_removed. At the centre of the cavity, the distance from the centre of the complete sphere is d = R/2. The field of the complete sphere there is E_complete = (Q / (4πε₀R³)) × (R/2) = Q / (8πε₀R²). The removed sphere is concentric with the cavity's centre, so at its own centre it produces zero electric field (E_removed = 0). Thus, E_net = Q / (8πε₀R²).

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4. A particle of mass m and carrying a charge -q moves in a circular path of radius r in the equatorial plane of an infinitely long, uniformly charged thin rod of linear charge density +λ. If the only force acting on the particle is electrostatic, its orbital time period T is proportional to:

  1. r^(1/2)
  2. r^(3/2)
  3. r
  4. Independent of r
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Answer: (C) r

The radial electrostatic force is F = q E = q λ / (2 π ε₀ r). Setting this equal to the centripetal force m v² / r, we get m v² / r = q λ / (2 π ε₀ r), which yields v² = q λ / (2 π ε₀ m) = constant. Since orbital speed v is independent of r, the period T = 2 π r / v is directly proportional to r.

easy

5. An electric dipole of dipole moment p is placed at the centre of a closed spherical surface of radius R. If the radius of the sphere is doubled, the electric flux through the surface:

  1. Is halved
  2. Remains zero
  3. Is doubled
  4. Becomes four times its initial value
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Answer: (B) Remains zero

An electric dipole consists of two equal and opposite charges (+q and -q). The net charge enclosed by the spherical Gaussian surface is q_enclosed = +q - q = 0. By Gauss's law, Φ = q_enclosed / ε₀ = 0, and doubling the radius keeps the enclosed charge zero, so flux remains zero.

hard

6. An infinitely long line of uniform linear charge density λ passes through a cube of edge length a. The maximum possible electric flux that can pass through the surface of the cube is:

  1. √3 λ a / ε₀
  2. √2 λ a / ε₀
  3. λ a / ε₀
  4. 2 λ a / ε₀
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Answer: (A) √3 λ a / ε₀

The electric flux through the cube is maximum when the length of the line enclosed by the cube is maximum. The longest line segment inside a cube of edge a is its body diagonal, which has length L_max = √3 a. Therefore, Φ_max = q_enc(max) / ε₀ = √3 λ a / ε₀.

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7. A closed cylindrical Gaussian surface of radius R and length L is aligned along the x-axis with its flat circular ends at x = 0 and x = L. An external non-uniform electric field in the region is given by E→ = (α x² î) N/C, where α is a positive constant. What is the net electric charge enclosed inside the cylinder?

  1. 2 ε₀ α π R² L
  2. ε₀ α π R² L²
  3. ε₀ α π R L²
  4. ε₀ α π R² L
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Answer: (B) ε₀ α π R² L²

The curved surface has normal perpendicular to î, so flux through it is zero. At x = 0, E = 0, so flux through the left face is zero. At x = L, E = α L² î and the outward area vector is A→ = π R² î, giving net flux Φ = α L² π R². By Gauss's law, Q_enclosed = ε₀ Φ = ε₀ α π R² L².

easy

8. A point charge q is placed symmetrically at the midpoint of one of the 12 edges of a cube. What is the total electric flux passing through the entire surface of this cube?

  1. q / (2ε₀)
  2. q / (8ε₀)
  3. q / (4ε₀)
  4. q / (12ε₀)
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Answer: (C) q / (4ε₀)

An edge of a cube is shared equally by 4 adjacent identical cubes. By symmetry, the total flux q/ε₀ from the charge is distributed equally among the 4 cubes, so the flux through one cube is q/(4ε₀).

medium

9. Two infinite, parallel, non-conducting thin sheets 1 and 2 have uniform surface charge densities +2σ and -σ, respectively. The magnitude of the electric field in the region between the two sheets is:

  1. σ / (2ε₀)
  2. σ / ε₀
  3. 3σ / (2ε₀)
  4. 3σ / ε₀
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Answer: (C) 3σ / (2ε₀)

In the region between the sheets, the field due to sheet 1 points away from it (magnitude 2σ/(2ε₀)) and the field due to sheet 2 points toward it (magnitude σ/(2ε₀)). Since both point in the same direction, the net field is (2σ + σ) / (2ε₀) = 3σ / (2ε₀).

medium

10. An electric field in a region is given by E→ = (3x² î) N/C. The electric flux through a cube bounded by the planes x = 0, x = 2 m, y = 0, y = 2 m, z = 0, and z = 2 m is:

  1. 0
  2. 24 N·m²/C
  3. 48 N·m²/C
  4. 96 N·m²/C
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Answer: (C) 48 N·m²/C

The field has only an x-component, so flux through faces parallel to the xy and xz planes is zero. At x = 0, E = 0, giving Φ₁ = 0. At x = 2 m, E = 3(2)² = 12 N/C pointing along +î. The area of the face is A = 2 × 2 = 4 m². Therefore, total flux Φ = E × A = 12 × 4 = 48 N·m²/C.

hard

11. The electric field in a certain region of space is directed radially outward from the origin with magnitude E(r) = α r², where α is a constant and r is the distance from the origin. The total charge enclosed within a sphere of radius R centered at the origin is:

  1. 4π ε₀ α R³
  2. 2π ε₀ α R³
  3. (4/3) π ε₀ α R⁴
  4. 4π ε₀ α R⁴
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Answer: (D) 4π ε₀ α R⁴

By Gauss's law, the total outward electric flux through the spherical surface of radius R is Φ = E(R) · (4π R²) = (α R²) (4π R²) = 4π α R⁴. Since Φ = q_enc / ε₀, the enclosed charge is q_enc = 4π ε₀ α R⁴.

hard

12. A spherical non-conducting shell of inner radius R₁ and outer radius R₂ has a uniform volume charge density ρ. What is the magnitude of the electric field at a distance r from the center, where R₁ < r < R₂?

  1. (ρ / 2ε₀) [ (r³ - R₁³) / r² ]
  2. (ρ / 3ε₀) [ (r² - R₁²) / r ]
  3. (ρ / ε₀) [ (r³ - R₁³) / R₂² ]
  4. (ρ / 3ε₀) [ (r³ - R₁³) / r² ]
Show answer and explanation

Answer: (D) (ρ / 3ε₀) [ (r³ - R₁³) / r² ]

Take a concentric spherical Gaussian surface of radius r. The enclosed charge is q_enc = ρ · (4/3)π (r³ - R₁³). Applying Gauss's law, E(4π r²) = q_enc / ε₀ = [4π ρ (r³ - R₁³)] / (3ε₀), which gives E = (ρ / 3ε₀) [ (r³ - R₁³) / r² ].

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