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Electric Flux and Gauss's Law

Electric flux through a surface measures the total electric field passing normally through that surface. For a closed surface, Gauss's law states that the net electric flux equals the enclosed charge divided by ε0: Φ = qenclosed/ε0.

In one line: Gauss's law counts charge inside the closed surface, not every charge nearby.

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Understand it

  • For a small area element dA, the area vector is perpendicular to the surface.
  • Electric flux is dΦ = E·dA = E dA cosθ.
  • For a closed surface, outward area vectors are used.
  • Gauss's law relates net flux only to charge enclosed by the closed surface, not to charges outside it.
  • It is especially useful when symmetry allows E to be constant over suitable parts of a Gaussian surface.

Example

If a closed surface encloses charge +2 microC, net flux through it is Φ = q/ε0 = (2 × 10−6)/(8.85 × 10−12) = 2.26 × 105 N m2 C−1 outward.

Formula and rule

Electric flux: Φ = ∫E·dA, SI unit N m2 C−1.

For uniform field over flat area, Φ = EA cosθ.

Gauss's law: ∮E·dA = qenclosed/ε0.

Use Gauss's law for closed surfaces; it becomes a practical field-finding tool only with high symmetry such as spherical, cylindrical, or planar symmetry.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A cube encloses a point charge of +8.85 nC.

  1. 1q = 8.85 × 10−9 C.
  2. 2Net flux Φ = q/ε0 = (8.85 × 10−9)/(8.85 × 10−12) = 1.0 × 103 N m2 C−1.
  3. 3This is total flux through all six faces together.

Method you can reuse

  1. 1Check whether the surface is open or closed.
  2. 2For open surface use Φ = EA cosθ when field is uniform.
  3. 3For closed surface identify enclosed charge.
  4. 4Apply ∮E·dA = qenclosed/ε0.
  5. 5Use symmetry only when E has constant magnitude and known direction over chosen surface parts.

Visual tip

Required diagrams: open flat surface with area vector, closed Gaussian surface with outward dA vectors, charge inside, charge outside, and field lines crossing the surface.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Does a charge outside a closed Gaussian surface contribute to net electric flux through that surface?

Think first, then show answer
No. An outside charge may create electric field on the surface, but the field lines entering and leaving the closed surface balance, so its contribution to net flux is zero.

Common wrong answer

Dividing total flux of a cube equally among faces without checking charge position or symmetry can be wrong. Equal sharing is valid only for a charge at the centre of a cube by symmetry.

Don't mix this up

Students often include charges outside the closed surface in qenclosed. External charges may affect field at points, but their net flux through the closed surface is zero.

Exam tip

Before applying Gauss's law, ask two things: what charge is enclosed, and whether symmetry lets you take E outside the integral.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Asked as statement of Gauss's law, flux calculation through a surface, reasoning about enclosed versus external charge, and derivations of fields using symmetric Gaussian surfaces.
  1. 1

    Enough for 1 mark

    Electric flux through a surface measures the total electric field passing normally through that surface. For a closed surface, Gauss's law states that the net electric flux equals the enclosed charge divided by ε0: Φ = qenclosed/ε0.

  2. 2

    Add for 2 marks

    Example: If a closed surface encloses charge +2 microC, net flux through it is Φ = q/ε0 = (2 × 10−6)/(8.85 × 10−12) = 2.26 × 105 N m2 C−1 outward.

  3. 3

    Add for 3 marks

    For a small area element dA, the area vector is perpendicular to the surface.

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