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Gauss's law and its applications

Learn Gauss's law, electric flux, and how to use symmetry to find electric fields of charged spheres, shells, and other symmetric objects.

Why this shows up in the exam

NEET often asks you to use Gauss's law for quick field calculations in symmetric situations.

How NEET tests this

Numerical · 2 QsDirect recall · 2 QsApplication

Learn the idea

Gauss's law states that the total electric flux through any closed surface equals the enclosed charge divided by ε₀; the key insight is that a zero net flux means the inward and outward flux lines cancel, not that the field must vanish.

🧠 Memory hook: Zero flux is like a roundabout: as many cars go in as come out – traffic moves but the net count is zero.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Electric flux Φ = ∮E·dA
  • Gauss's law: ∮E·dA = Q_enclosed/ε₀
  • For a uniformly charged spherical shell, E = 0 inside and E = (1/4πɛ₀) Q/r² outside
  • For a uniformly charged solid sphere, E = (1/4πɛ₀) Q r /R³ inside (field varies linearly)
  • Choose a Gaussian surface so that E is constant on each part and normal to the surface
  • ε₀ = 8.85×10⁻¹² F·m⁻¹

How to approach it

  1. 1Identify the symmetry of the charge distribution (spherical, cylindrical, planar)
  2. 2Draw a Gaussian surface that matches that symmetry
  3. 3Write the flux integral; replace E by its constant value and evaluate the area factor
  4. 4Set the integral equal to Q_enclosed/ε₀ and solve for the magnitude of E

Worked example — watch it click

If ∮⃗E⃗⋅d⃗s⃗ = 0 over a surface, then

  • A)the magnitude of the electric field on the surface is constant.
  • B)all the charges must necessarily be inside the surface.
  • C)the electric field inside the surface is necessarily uniform.
  • ✅the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

The concept behind this problem

The worked example tests whether you interpret a zero Gauss integral as a balance of inward and outward flux lines, not as a statement about field uniformity or charge location.

Step by step

  1. 1By Gauss's law, ∮E⃗·ds⃗ = q_enclosed/ε₀.
  2. 2If this integral equals zero, then q_enclosed = 0, meaning net charge inside is zero (could have +q and -q, or no charges at all).
  3. 3This means net flux through surface is zero, so flux entering equals flux leaving.
  4. 4Option (a) is false - E magnitude need not be constant.
  5. 5Option (b) is false - charges can be outside too.
  6. 6Option (c) is false - E need not be uniform.
  7. 7Option (d) correctly states that net flux = 0 means inward flux = outward flux.

Watch out

Students often think zero flux means the electric field must be constant (or zero) on the surface, which is incorrect.

Common slip-ups that cost marks

  • •Assuming zero net flux forces E to be zero everywhere on the surface
  • •Using an open surface or a surface that does not enclose the charge
  • •Forgetting that charges outside the Gaussian surface do not contribute to Q_enclosed

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 5NEET 2023

If ∮⃗E⃗⋅d⃗s⃗ = 0 over a surface, then

Push further

More challenging

9 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 9

A hollow conducting sphere of radius R has a charge +Q placed at its center. An additional charge +2Q is given to the conductor itself. What is the electric flux through a concentric spherical Gaussian surface of radius 2R?

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