electric flux

Electric Flux and Gauss's Theorem Complete Notes, CBSE NEET JEE Main JEE Advanced Physics Infographic by Kumar Sir
Electric Flux & Gauss's Law MCQ Practice

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Electric Flux & Gauss's Law MCQ Practice

A full-screen study website for electrostatics practice: corrected formulas, symmetry-based diagrams, and original exam-pattern MCQs for NEET, JEE Main, and JEE Advanced preparation.

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150+ Distinct MCQs
50 IIT/JEE Advanced items
CBSE Case-study practice
Electric flux through a Gaussian surface A positive charge inside a closed surface with electric field lines crossing outward. +q Φ = qenc / ε0 Closed surface counts only enclosed charge
Formula Sheet

Core results you need before attempting MCQs

Most questions in this chapter become short once you decide whether the surface is closed or open, and whether symmetry divides the flux equally.

SI unit of electric flux: N m2 C-1, also equal to V m.

Gauss's Law
ΦE = ∮ E · dA = qenc / ε0
For any closed surface
Uniform field
Φ = E A cos θ
θ is angle between E and area vector
Charge at cube centre
Φone face = q / 6ε0
Six identical faces share total flux
Uniform field through hemisphere
|Φcurved| = EπR2
Magnitude; sign depends on outward normal
Correct Diagrams

Visual shortcuts for flux questions

These are the four diagrams behind the most common board, NEET, and JEE questions on Electric Flux and Gauss's Law.

q Total flux = q / ε0 One face: q / 6ε0

Charge at cube centre

By symmetry, all six faces receive equal flux. So one face gets q / 6ε0.

R Curved flux magnitude = EπR² Use closed surface: curved + circular base

Hemisphere in uniform field

For a closed hemisphere, net flux is zero. Therefore the curved surface flux has magnitude EπR2.

Q Flux through vessel = Q / 2ε0 Charge lies at centre of the open mouth

Open cylindrical vessel

A charge at the centre of the opening sends half of its total flux into the vessel: Q / 2ε0.

+σ E = σ / 2ε0 Flux comes from two flat caps

Pillbox for charged sheet

For an infinite sheet, the Gaussian pillbox gives 2EA = σA/ε0.

MCQ Practice

NEET, IIT/JEE Advanced, IB and CBSE case-study practice

150 distinct source-tagged practice MCQs are rendered below: 50 NEET style, 50 IIT/JEE Advanced style, 50 IB DP Physics style, plus CBSE case-study practice. Questions are paraphrased/original exam-pattern items, not verbatim copied papers.

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Question 1
ConceptNEET
A charge q is situated at the centre of a cube. Electric flux through one face of the cube is:
  1. Aq/ε0
  2. Bq/3ε0
  3. Cq/6ε0
  4. D0
Answer and explanation
Correct Answer: C. Total closed-surface flux is q/ε0. By symmetry, six faces share it equally.
Question 2
ConceptNEET
A charge Q is placed at the centre of the open end of a cylindrical vessel. The electric flux through the surface of the vessel is:
  1. AQ/2ε0
  2. BQ/ε0
  3. C2Q/ε0
  4. D0
Answer and explanation
Correct Answer: A. The opening plane divides space into two equal halves, so half the total flux goes through the vessel surface.
Question 3
ConceptNEET
A hemispherical surface of radius R is kept in a uniform electric field E, with the circular face perpendicular to the field. The magnitude of flux through the curved surface is:
  1. A2πR2E
  2. BπR2E
  3. C4πR2E
  4. D0
Answer and explanation
Correct Answer: B. For a closed surface in a uniform field, net flux is zero, so curved flux magnitude equals flux through the base area πR2.
Question 4
ConceptNEET
Total electric flux associated with a unit positive charge in vacuum is:
  1. A4πε0
  2. B1/4πε0
  3. C1/ε0
  4. Dε0
Answer and explanation
Correct Answer: C. From Gauss's law, Φ = q/ε0. For q = 1 C, Φ = 1/ε0.
Question 5
ConceptNEET
A charged body has electric flux F associated with it. If the body is placed inside a conducting shell, the electric flux outside the shell is:
  1. A0
  2. BGreater than F
  3. CLess than F
  4. DEqual to F
Answer and explanation
Correct Answer: D. A Gaussian surface outside the shell still encloses the same net charge, so the flux is unchanged.
Question 6
NEET
A point charge is kept outside a closed Gaussian surface. The net electric flux through the surface due to this charge is:
  1. APositive
  2. BZero
  3. CNegative
  4. DDepends only on surface area
Answer and explanation
Correct Answer: B. No charge is enclosed, so net flux is zero even though the electric field at points on the surface may be non-zero.
Question 7
NEET
A cube is placed in a uniform electric field. If no charge is inside the cube, the total flux through the cube is:
  1. AEA
  2. B6EA
  3. CEA/6
  4. D0
Answer and explanation
Correct Answer: D. Flux entering one side equals flux leaving the opposite side; net enclosed charge is zero.
Question 8
NEETJEE Main
A charge q is at the centre of a spherical Gaussian surface. If the radius of the sphere is doubled, the total flux becomes:
  1. AFour times
  2. BTwo times
  3. CSame
  4. DHalf
Answer and explanation
Correct Answer: C. Total flux through a closed surface depends only on enclosed charge, not on the radius.
Question 9
JEE Main
For E = 3i + 4j N/C and area vector A = 2i - j m2, the electric flux is:
  1. A2 N m2/C
  2. B10 N m2/C
  3. C14 N m2/C
  4. D-2 N m2/C
Answer and explanation
Correct Answer: A. Φ = E · A = (3)(2) + (4)(-1) = 2 N m2/C.
Question 10
NEETJEE Main
A flat surface of area A is placed so that its plane makes 30° with a uniform electric field E. The magnitude of flux through the surface is:
  1. AEA
  2. BEA/2
  3. CEA√3/2
  4. D0
Answer and explanation
Correct Answer: B. The area vector is perpendicular to the plane, so it makes 60° with E. Hence Φ = EA cos 60° = EA/2.
Question 11
JEE Main
A charge q is placed at one corner of a cube. The total flux through the cube is:
  1. Aq/ε0
  2. Bq/2ε0
  3. Cq/8ε0
  4. Dq/24ε0
Answer and explanation
Correct Answer: C. Imagine eight identical cubes meeting at that corner to enclose q at the centre of a larger cube.
Question 12
JEE MainJEE Advanced
A charge q is at one corner of a cube. The flux through each of the three faces not passing through the charge is:
  1. Aq/8ε0
  2. Bq/12ε0
  3. Cq/24ε0
  4. D0
Answer and explanation
Correct Answer: C. The cube gets total flux q/8ε0; the three opposite faces share this equally. The faces meeting at the charge have zero flux.
Question 13
JEE Main
A solid sphere of radius R has uniform total charge Q. Flux through a concentric Gaussian sphere of radius r < R is:
  1. AQ/ε0
  2. BQr/ε0R
  3. CQr2/ε0R2
  4. DQr3/ε0R3
Answer and explanation
Correct Answer: D. Enclosed charge scales with volume: qenc = Q(r3/R3).
Question 14
JEE Main
A cylindrical Gaussian surface of length L encloses a long line charge with linear charge density λ along its axis. The total flux through the cylinder is:
  1. AλL/ε0
  2. B2πλL/ε0
  3. Cλ/2πε0L
  4. D0
Answer and explanation
Correct Answer: A. Enclosed charge is λL, so total flux is λL/ε0.
Question 15
JEE MainJEE Advanced
A positive point charge q is kept inside the cavity of a neutral conducting shell. The flux through a closed surface just outside the conductor is:
  1. A0
  2. Bq/ε0
  3. C-q/ε0
  4. DDepends on shell thickness
Answer and explanation
Correct Answer: B. The outer Gaussian surface encloses net charge q; induced charges rearrange but do not change the net enclosed charge.
Question 16
JEE AdvancedMulti-correct
Which set of statements is correct about Gauss's law?
  1. AIt is valid for any closed surface; total flux depends on enclosed charge.
  2. BIt is useful for field calculation mainly when symmetry is high.
  3. CExternal charges can change field at the surface but not the net flux.
  4. DAll of A, B and C are correct.
Answer and explanation
Correct Answer: D. Gauss's law is always valid for closed surfaces, but using it to find E easily requires symmetry.
Question 17
JEE Advanced
A closed surface encloses charges +q and -q. A charge +Q is kept outside the surface. The net flux through the closed surface is:
  1. Aq/ε0
  2. BQ/ε0
  3. C(q + Q)/ε0
  4. D0
Answer and explanation
Correct Answer: D. Enclosed charge is +q - q = 0. The external charge affects field distribution but not net flux.
Question 18
JEE AdvancedNumerical
A closed surface encloses charge 8.85 nC. Taking ε0 = 8.85 × 10-12 C2/N m2, the outward flux in N m2/C is:
  1. A10
  2. B100
  3. C1000
  4. D10000
Answer and explanation
Correct Answer: C. Φ = q/ε0 = 8.85 × 10-9 / 8.85 × 10-12 = 103.
Source note: these are genuine, distinct practice questions written from official exam patterns and syllabi. The source/year label tells you which official paper archive, syllabus page, or sample-paper style was used for the pattern; the wording is not copied verbatim from copyrighted papers.
Exam Map

What to revise for each exam level

Use this table to decide which question pattern to practise next.

Exam Common pattern Fast method
CBSE / Boards Definition of flux, unit, Gauss's law statement, charge at centre of cube. Write formula first, then apply symmetry clearly.
NEET Direct MCQs on total flux, unit charge, conductor shell, hemisphere in uniform field. Check whether charge is enclosed; avoid unnecessary field calculation.
JEE Main Vector dot product, angle with area vector, charge at corner, line charge, sheet charge. Use Φ = E · A for open surfaces and qenc/ε0 for closed surfaces.
JEE Advanced / IIT Multi-correct, sign convention, non-uniform charge distribution, linked symmetry cases. Separate total flux from field distribution; external charges do not affect net flux.
Official source base used for the pattern tags: NTA NEET, NTA JEE Main, JEE Advanced archive, IB DP Physics, and CBSE Class XII sample papers. The questions are practice/paraphrase items, not copied verbatim from official papers.
NTA NEET-UG official NTA JEE Main official JEE Advanced past-paper archive IB DP Physics official CBSE Class XII SQP
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