Gauss's Law says the total electric flux through any closed surface equals the net charge enclosed inside it divided by epsilon-0: Phi = q_enclosed / epsilon-0. Only the charge INSIDE the surface matters; the shape, size, or outside charges do not change the total flux. Memory hook: "flux counts only the charge trapped in the bag."
Left: a single +q inside a closed surface gives total flux q/epsilon-0 (lines point outward). Right: a dipole (+q and -q) inside gives net charge zero, so total flux is zero even though the field is not zero.
Your doubts, answered
Does the flux depend on the shape or size of the surface?
No. Gauss's Law gives Phi = q_enclosed / epsilon-0. The total flux depends ONLY on the net charge trapped inside and the constant epsilon-0. Whether you draw a cube, a sphere, or a lumpy blob around the same charge, the total flux is the same. Shape, size, area, and volume do not appear in the formula. This exact idea was asked in NEET 2023.
What does 'q enclosed' actually mean?
q_enclosed is the NET charge sitting inside the closed surface (add signs). If a surface holds +5 C and -2 C, then q_enclosed = +3 C. Charges lying outside the surface are ignored for the TOTAL flux, because every field line from an outside charge that enters the surface also leaves it, giving zero net contribution.
Why is the flux through a surface enclosing a dipole zero?
A dipole is +q and -q together. The net charge inside is (+q) + (-q) = 0. So Phi = q_enclosed / epsilon-0 = 0 / epsilon-0 = 0. The field is NOT zero anywhere, but the number of lines going out equals the number coming in, so the net flux is zero. NEET 2019 (Odisha) tested this directly.
Do charges outside the surface change the flux?
They change the electric field E at points on the surface, but they do NOT change the total flux. Each outside charge sends the same number of lines in as out. So in Phi = q_enclosed / epsilon-0, you plug in only the inside charge. This is why zero flux does not mean zero field.
Is Gauss's Law true only for spheres and symmetric shapes?
Gauss's Law is ALWAYS true for any closed surface and any charge arrangement. Symmetry is only needed when you want to pull E out of the integral to CALCULATE the field (line, sheet, shell). The law itself Phi = q_enclosed / epsilon-0 holds universally.
⚠️ The NEET trap ✗ Thinking the flux gets bigger if you draw a bigger surface, or that a cube gives different flux than a sphere around the same charge. ✓ For a fixed enclosed charge, the total flux is fixed at q_enclosed / epsilon-0 no matter the surface's shape or size. A charge Q at a cube's centre gives Q/epsilon-0 total, so each of the 6 faces gets Q/(6 epsilon-0). 🧠 Bigger bag, same charge inside = same total flux. The bag's shape never changes the count.
Real NEET questions
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: Phi = q_enclosed / epsilon-0. Step 1: The right side contains only q_enclosed and the constant epsilon-0. Step 2: Shape, area, and volume do NOT appear anywhere in the formula. Step 3: So the total flux depends only on the net charge enclosed. Answer: option D.
2019
A sphere encloses an electric dipole with charges +/- 3 x 10^-6 C. What is the total electric flux across the sphere?
A · -3 x 10^-6 N m^2/C
B · zero ✓
C · 3 x 10^-6 N m^2/C
D · 6 x 10^-6 N m^2/C
Solution: Step 1: Gauss's Law gives Phi = q_enclosed / epsilon-0. Step 2: A dipole is +q and -q, so net charge inside = (+3 x 10^-6) + (-3 x 10^-6) = 0. Step 3: Phi = 0 / epsilon-0 = 0. The field is not zero, but the net flux is zero. Answer: option B.
2023
A charge Q uC 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 x 10^-3) / (6 epsilon-0)
B · (Q x 10^-6) / (6 epsilon-0) ✓
C · (Q x 10^-6) / epsilon-0
D · (3Q x 10^-3) / (2 epsilon-0)
Solution: Step 1: Convert charge: Q uC = Q x 10^-6 C. Step 2: Total flux through the whole cube = q_enclosed / epsilon-0 = (Q x 10^-6) / epsilon-0. Step 3: By symmetry (charge at centre), the 6 faces share it equally, so one face gets (1/6) of the total = (Q x 10^-6) / (6 epsilon-0). Answer: 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.
Phi = closed integral of E . dS = q_enclosed / epsilon-0. In words, the total electric flux through a closed surface equals the net enclosed charge divided by the permittivity of free space epsilon-0 (about 8.85 x 10^-12 C^2 N^-1 m^-2).
What is the SI unit of electric flux?
The SI unit of electric flux is N m^2/C (newton metre squared per coulomb), which is the same as volt-metre (V m).
Is Gauss's Law valid for a non-uniform or asymmetric charge?
Yes. The law is always valid for any closed surface. Symmetry is only required when you use it to compute the field E, not for the law itself.
Why does epsilon-0 appear in Gauss's Law?
epsilon-0 (permittivity of free space) comes from Coulomb's law, since the flux from a point charge is derived using E = q / (4 pi epsilon-0 r^2). The 4 pi r^2 (the sphere's area) cancels, leaving q / epsilon-0.
Does zero flux mean zero electric field?
No. Zero net flux only means the net enclosed charge is zero (lines in = lines out). The field can still be strong at points on the surface, for example around a dipole.
