Flux Through a Cube and a Sphere (Enclosed Charge)
Physics · Electric Charges And Fields · NEET
By Gauss's law, the total electric flux through ANY closed surface (cube or sphere) is Φ = q_enclosed / ε₀. It does not care about the shape or size of the surface, only the net charge trapped inside. Memory hook: "Flux counts the charge, not the box." So a charge q at the centre of a cube gives total flux q/ε₀, and each of the 6 faces gets q/6ε₀.
A charge q enclosed by a cube or a sphere gives the same total flux Φ = q/ε₀, because Gauss's law depends only on the enclosed charge, not the shape. For a charge at the centre of a cube, each of the 6 faces carries an equal share q/6ε₀.
Your doubts, answered
Why is the flux through a cube the same as through a sphere for the same charge?
Gauss's law says Φ = q_enclosed / ε₀. The right side has ONLY the enclosed charge and ε₀ — no term for shape, size, or the distance of the surface. So if a cube and a sphere both enclose the same charge q, both have exactly the same total flux q/ε₀. The field lines that start on the charge must all pass out through whatever closed surface surrounds it, whether it is round or square.
How do I find flux through just ONE face of a cube?
First find the TOTAL flux, then use symmetry. If charge q sits at the CENTRE of the cube, all 6 faces are identical by symmetry, so each face carries an equal share: flux per face = (1/6) × (q/ε₀) = q/6ε₀. This equal-sharing trick only works when the charge is at the centre. If the charge is not centred, the faces are no longer equal and you cannot just divide by 6.
What if the charge is at a CORNER of the cube instead of the centre?
A charge at a corner is shared by 8 cubes meeting at that corner, so only 1/8 of its flux enters our cube: total flux = q/8ε₀. Of the 3 faces touching the corner, the field is parallel to them, so they get zero flux. The other 3 faces share the q/8ε₀ equally, giving q/24ε₀ per face. This is a favourite NEET twist on the centre case.
Does the flux change if I make the sphere bigger?
No. As long as the same charge stays enclosed, growing the sphere does NOT change the total flux — it stays q/ε₀. A bigger sphere has a weaker field E (since E ∝ 1/r²) but a larger area (4πr²), and the two effects cancel exactly. Flux depends on enclosed charge alone, never on radius.
What is the flux if the charge is OUTSIDE the closed surface?
Zero. Every field line from an outside charge that enters the surface must also leave it, so inward flux cancels outward flux. Since q_enclosed = 0, Gauss's law gives Φ = 0. Remember: only charges INSIDE the surface contribute to net flux; outside charges affect the field on the surface but not the net flux.
⚠️ The NEET trap ✗ Charge q at the centre of a cube, so flux through the whole cube = q/6ε₀. ✓ q/6ε₀ is the flux through ONE face. The TOTAL flux through the whole cube is q/ε₀ (all 6 faces together). 🧠 Read whether the question asks 'one face' or 'whole cube'. Total = q/ε₀; per face (centre only) = q/6ε₀.
Real NEET questions
2023
A charge Q μC is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit):
A · (Q × 10⁻³) / (6ε₀)
B · (Q × 10⁻⁶) / (6ε₀) ✓
C · (Q × 10⁻⁶) / ε₀
D · (3Q × 10⁻³) / (2ε₀)
Solution: Step 1: Convert units. Charge = Q μC = Q × 10⁻⁶ C.
Step 2: Total flux through the closed cube (Gauss's law): Φ_total = q_enclosed / ε₀ = (Q × 10⁻⁶) / ε₀.
Step 3: Charge is at the CENTRE, so by symmetry all 6 faces get an equal share.
Flux per face = (1/6) × Φ_total = (Q × 10⁻⁶) / (6ε₀).
So the answer is option B.
2023
According to Gauss's law of electrostatics, the electric flux through a closed surface depends on:
A · the shape of the surface
B · the volume enclosed by the surface
C · the area of the surface
D · the quantity of charge enclosed by the surface ✓
Solution: Gauss's law: Φ = ∮ E · dS = q_enclosed / ε₀.
The right-hand side contains ONLY the enclosed charge and ε₀. There is no term for shape, area, or volume.
So the net flux through any closed surface (cube, sphere, any shape) depends only on the total charge enclosed. Answer: option D.
2019
A sphere encloses an electric dipole with charges ±3 × 10⁻⁶ C. What is the total electric flux across the sphere?
A · −3 × 10⁻⁶ N m²/C
B · zero ✓
C · 3 × 10⁻⁶ N m²/C
D · 6 × 10⁻⁶ N m²/C
Solution: Gauss's law: Φ = q_enclosed / ε₀.
A dipole has equal and opposite charges +q and −q.
Net enclosed charge = (+3 × 10⁻⁶) + (−3 × 10⁻⁶) = 0.
Therefore Φ = 0 / ε₀ = 0. The flux is zero — option B.
Solved Electric Charges And Fields NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for flux through a cube with a charge at its centre?
Total flux through the cube = q/ε₀. Flux through each one of the 6 faces = q/6ε₀, because a centred charge shares its flux equally among all faces.
Is flux through a sphere different from flux through a cube?
No, not the total flux. For the same enclosed charge, both give q/ε₀. Only the field distribution differs; the net flux is identical because Gauss's law ignores shape.
Does flux depend on where the charge is placed inside the surface?
The TOTAL flux does not — it is q/ε₀ wherever the charge sits inside. But the flux through individual faces DOES change; equal sharing (q/6ε₀ per face) only holds when the charge is at the centre.
What is the SI unit of electric flux?
Electric flux is measured in N·m²/C (newton metre squared per coulomb), which is the same as volt·metre (V·m).
If I double the charge inside a sphere, what happens to the flux?
The flux doubles. Since Φ = q/ε₀, flux is directly proportional to the enclosed charge. Doubling q doubles Φ, no matter the size or shape of the surface.
