Continuous Charge Distribution, Electric Flux and Gauss's Law – continuous charge distribution: NEET MCQs with Answers

15 free questions on Continuous Charge Distribution, Electric Flux and Gauss's Law – continuous charge distribution for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: continuous charge distribution, Electric flux. Gauss's law and its applications to find field due to infinitely long uniformly charged straight wire, uniformly charged infinite plane sheet, and uniformly charged thin spherical shell.

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Question 1

Previous-year question
A spherical conductor of radius 10 cm has a charge of 3.2 × 10⁻⁷ C distributed uniformly. What is the magnitude of the electric field at a point 15 cm from the centre of the sphere? [NEET 2020]
  1. 1.28 × 10⁵ N C⁻¹
  2. 1.28 × 10⁶ N C⁻¹
  3. 1.28 × 10⁷ N C⁻¹
  4. 1.28 × 10⁴ N C⁻¹
Show answer
Answer: (A) 1.28 × 10⁵ N C⁻¹
Explanation: Outside the conductor it behaves as a point charge: E = 9 × 10⁹ × 3.2 × 10⁻⁷/(0.15)² = 1.28 × 10⁵ N C⁻¹.

Question 2

Previous-year question
Two parallel infinite line charges with linear charge densities +λ C m⁻¹ and −λ C m⁻¹ are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges? [NEET 2019]
  1. zero
  2. 2λ/πε₀R
  3. λ/πε₀R
  4. λ/2πε₀R
Show answer
Answer: (C) λ/πε₀R
Explanation: Each line gives λ/2πε₀R at the mid-point, and because the charges are of opposite sign both fields point the same way. Adding gives λ/πε₀R.

Question 3

A solid non-conducting sphere of radius a has a charge density ρ = kr, where k is a positive constant and r is the distance from the centre. If Q is the total charge of the sphere, which of the following gives Q and the electric field E at a point inside the sphere?
  1. Q = (4/3)πka⁴ and E = Qr/(4πε₀a³)
  2. Q = πka⁴ and E = Qr²/(4πε₀a⁴)
  3. Q = πka⁴ and E = Qr/(4πε₀a³)
  4. Q = (4/3)πka⁴ and E = Qr²/(4πε₀a⁴)
Show answer
Answer: (B) Q = πka⁴ and E = Qr²/(4πε₀a⁴)
Explanation: The charge within radius r is the integral of (kr)(4πr²)dr, which equals πkr⁴, so the total charge is Q = πka⁴. The charge enclosed at radius r is therefore Q(r⁴/a⁴). Gauss's law gives E(4πr²) = Qr⁴/(ε₀a⁴), so E = Qr²/(4πε₀a⁴).

Question 4

The electric field on the axis of a thin uniformly charged disc of radius R, at a distance z from its centre, is E = (σ/2ε₀)[1 − z/√(z² + R²)]. For a charged plastic disc the field is 2.0 × 10⁷ N C⁻¹ very close to the centre and 0.80 × 10⁷ N C⁻¹ at z = 9.0 cm. The radius and the total charge of the disc are (take ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
  1. R = 15 cm and q = 25 μC
  2. R = 6.0 cm and q = 4.0 μC
  3. R = 12 cm and q = 16 μC
  4. R = 12 cm and q = 8.0 μC
Show answer
Answer: (C) R = 12 cm and q = 16 μC
Explanation: Close to the centre E = σ/2ε₀, so σ = 2ε₀(2.0 × 10⁷) = 3.54 × 10⁻⁴ C m⁻². The ratio of the two readings gives 1 − z/√(z² + R²) = 0.40, so √(z² + R²) = z/0.60 = 15 cm and R = 12 cm. The charge is then q = σπR² ≈ 1.6 × 10⁻⁵ C.

Question 5

A thin conducting spherical shell of radius R carries a charge Q. How does the magnitude of the electric field E vary with the distance r from its centre?
  1. E rises linearly with r at all distances
  2. E is constant and non-zero for r R
  3. E is zero for r < R, is largest just outside the surface, and falls as 1/r² for r > R
  4. E rises linearly with r up to r = R and then falls as 1/r²
Show answer
Answer: (C) E is zero for r < R, is largest just outside the surface, and falls as 1/r² for r > R
Explanation: All the charge on a conductor lies on its surface, so a Gaussian sphere with r R the shell acts like a point charge Q at its centre, so E = Q/(4πε₀r²). The field is therefore greatest just outside the surface, at Q/(4πε₀R²), and falls as 1/r² beyond it.

Question 6

A thin non-conducting sheet and the flat outer surface of a charged metal block have the same uniform surface charge density σ. Close to each surface, how does the field near the metal compare with the field near the sheet, and why?
  1. Twice as large, because the field inside the metal is zero, so all the flux leaves through one side
  2. Equal, because Gauss's law gives the same flux for the same enclosed charge in both cases
  3. Twice as large, because a conductor holds twice as much charge in the same surface area
  4. Half as large, because the charges in a metal are free to move away from the surface
Show answer
Answer: (A) Twice as large, because the field inside the metal is zero, so all the flux leaves through one side
Explanation: For a thin sheet, a Gaussian pillbox has flux through both faces, so 2EA = σA/ε₀ and E = σ/2ε₀. For a conductor, the field inside the metal is zero, so flux passes only through the outer face. Then EA = σA/ε₀ and E = σ/ε₀, which is twice as large.

Question 7

Point A lies just outside the flat surface of a large charged conductor, where the surface charge density is σ. Point B lies close to a large thin non-conducting sheet that also has uniform surface charge density σ. What is the ratio of the electric field magnitude at A to that at B?
  1. 1 : 2
  2. 4 : 1
  3. 1 : 1
  4. 2 : 1
Show answer
Answer: (D) 2 : 1
Explanation: For the conductor the field inside is zero, so all the flux through a small Gaussian pillbox leaves through its outer face, giving E = σ/ε₀. For the thin sheet the flux leaves through both faces of the pillbox equally, giving E = σ/2ε₀. The ratio is therefore 2 : 1.
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Question 8

The electric field is measured at a distance r from each of four charged objects: (i) an infinite uniformly charged plane sheet, (ii) an infinitely long uniformly charged wire, (iii) a point charge and (iv) a short electric dipole, on its axis. The distance is then increased to 3r in each case, with r much larger than the dipole length. By what factors, in the order (i), (ii), (iii), (iv), does the field decrease?
  1. 1, 9, 3, 27
  2. 1, 3, 9, 27
  3. 3, 1, 9, 27
  4. 1, 3, 9, 9
Show answer
Answer: (B) 1, 3, 9, 27
Explanation: The field of an infinite sheet is σ/2ε₀, which does not depend on distance. A long wire gives E ∝ 1/r, a point charge gives E ∝ 1/r², and a short dipole gives E ∝ 1/r³ at large distances. Tripling r therefore divides the fields by 1, 3, 9 and 27.

Question 9

A large isolated thin metal plate of area A carries a total charge Q. A large thin non-conducting sheet of the same area carries the same total charge Q spread uniformly over it. What is the ratio of the electric field just outside the central part of the metal plate to that just outside the central part of the sheet?
  1. 1 : 1
  2. 4 : 1
  3. 1 : 2
  4. 2 : 1
Show answer
Answer: (A) 1 : 1
Explanation: On the metal plate the charge resides on the two faces, so each face has density Q/(2A). The field just outside a conductor is (surface density)/ε₀, which gives Q/(2Aε₀). For the non-conducting sheet, σ = Q/A and the field is σ/(2ε₀) = Q/(2Aε₀). The two fields are equal, although the formulas look different, because the conductor's σ refers to the charge on one face only.

Question 10

A closed surface encloses point charges +3 μC and −1 μC, while a charge of +5 μC lies outside it. In Gauss's law applied to this surface, which statement is correct?
  1. E at points on the surface is due to the two enclosed charges only, and the net flux is (2 μC)/ε₀
  2. E at points on the surface is due to all three charges, and the net flux is (7 μC)/ε₀
  3. E at points on the surface is due to the two enclosed charges only, and the net flux is (7 μC)/ε₀
  4. E at points on the surface is due to all three charges, and the net flux is (2 μC)/ε₀
Show answer
Answer: (D) E at points on the surface is due to all three charges, and the net flux is (2 μC)/ε₀
Explanation: The field E in the surface integral is the resultant field at each point, so the outside charge also contributes to it. Field lines from the outside charge enter and leave the closed surface equally, so that charge adds nothing to the net flux. The net flux therefore depends only on the enclosed charge, (+3 − 1) μC = 2 μC, divided by ε₀.

Question 11

Suppose the electric field of a point charge varied as 1/r³ instead of 1/r². If the flux through a sphere of radius R centred on the charge is φ, the flux through a concentric sphere of radius 2R would be
  1. φ/2
  2. φ
  3. φ/4
  4. 2φ
Show answer
Answer: (A) φ/2
Explanation: Flux through a concentric sphere is the field multiplied by 4πr². With E proportional to 1/r³, the flux is proportional to 1/r, so doubling the radius halves it. Gauss's law gives a flux independent of the surface only because the actual field follows an inverse-square law.

Question 12

A point charge lies on the axis of an imaginary circular disc, at a distance l from the centre of the disc. For exactly one-fourth of the total flux emitted by the charge to pass through the disc, the radius of the disc must be
  1. √3 l
  2. 2l
  3. l/√3
  4. l
Show answer
Answer: (A) √3 l
Explanation: If the disc subtends a half-angle θ at the charge, the fraction of the total flux through it is (1 − cos θ)/2. Setting this equal to 1/4 gives cos θ = 1/2, so θ = 60°. Then R = l tan 60° = √3 l.

Question 13

In a region the electric field is E = (200x) î N/C, where x is in metres. An imaginary cube of side 0.5 m has its edges parallel to the axes, with two faces lying in the planes x = 1.0 m and x = 1.5 m. The net charge enclosed by the cube is (take ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
  1. 4.4 × 10⁻¹⁰ C
  2. 1.1 × 10⁻¹⁰ C
  3. 6.6 × 10⁻¹⁰ C
  4. 2.2 × 10⁻¹⁰ C
Show answer
Answer: (D) 2.2 × 10⁻¹⁰ C
Explanation: The field is parallel to the other four faces, so only the faces at x = 1.0 m and x = 1.5 m carry flux. These fluxes are −200 × 0.25 = −50 and +300 × 0.25 = +75 N m²/C, giving a net flux of 25 N m²/C. By Gauss's law, q = ε₀φ = 8.85 × 10⁻¹² × 25 ≈ 2.2 × 10⁻¹⁰ C.

Question 14

A solid non-conducting sphere with centre O has uniform positive volume charge density ρ. A spherical cavity with centre C is hollowed out wholly inside it, with OC = d. The electric field at any point inside the cavity is
  1. uniform, of magnitude ρd/ε₀, directed from O towards C
  2. uniform, of magnitude ρd/(3ε₀), directed from O towards C
  3. zero everywhere in the cavity
  4. uniform, of magnitude ρd/(3ε₀), directed from C towards O
Show answer
Answer: (B) uniform, of magnitude ρd/(3ε₀), directed from O towards C
Explanation: Treat the cavity as the full sphere of density +ρ together with a smaller sphere of density −ρ. Inside a uniformly charged sphere the field is ρr/(3ε₀) directed radially, so the two fields at a point P are (ρ/3ε₀)OP and −(ρ/3ε₀)CP. Their sum is (ρ/3ε₀)OC, which does not depend on the position of P.

Question 15

A thin fixed ring of radius R carries a uniformly distributed charge +Q. A particle of charge −q is released from rest on the axis of the ring, very close to its centre, and oscillates with time period T. If the charge on the ring is doubled and its radius is made four times, the time period of the same particle (still released very close to the centre) becomes
  1. 2√2 T
  2. 8T
  3. 4√2 T
  4. 16T
Show answer
Answer: (C) 4√2 T
Explanation: For x much less than R, the field on the axis is kQx/R³, so the restoring force on the particle is kQq x/R³. This gives T = 2π√(mR³/(kQq)), so T is proportional to √(R³/Q). The new period is T√(64/2) = √32 T = 4√2 T.

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