C
CraftExam
high importancemedium8 min

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 = q_enclosed/ε0.

Concept Practice Coming Soon

Learn the concept

Student-friendly explanation

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

How to write this in exams

  1. 1

    Start with the exact idea

    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 = q_enclosed/ε0.

  2. 2

    Then show how to use it

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

  3. 3

    Add one concrete example

    If a closed surface encloses charge +2 microC, net flux through it is Φ = q/ε0 = (2 x 10^-6)/(8.85 x 10^-12) = 2.26 x 10^5 N m^2 C^-1 outward.

  4. 4

    Avoid this incomplete 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.

Definition

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 = q_enclosed/ε0.

Example

If a closed surface encloses charge +2 microC, net flux through it is Φ = q/ε0 = (2 x 10^-6)/(8.85 x 10^-12) = 2.26 x 10^5 N m^2 C^-1 outward.

Rule to remember

Electric flux: Φ = ∫E·dA, SI unit N m^2 C^-1. For uniform field over flat area, Φ = EA cosθ. Gauss's law: ∮E·dA = q_enclosed/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.

Memory hook

Gauss's law counts charge inside the closed surface, not every charge nearby.

Examples and method

Worked example

A cube encloses a point charge of +8.85 nC. q = 8.85 x 10^-9 C. Net flux Φ = q/ε0 = (8.85 x 10^-9)/(8.85 x 10^-12) = 1.0 x 10^3 N m^2 C^-1. This is total flux through all six faces together.

Method to apply

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

Diagram support

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.

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.

Avoid common mistakes

Common confusion

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

Exam tip

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

Quick check

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

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.

Answer writing and exam use

1-mark answer

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 = q_enclosed/ε0.

2-mark answer

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 = q_enclosed/ε0. Electric flux: Φ = ∫E·dA, SI unit N m^2 C^-1. For uniform field over flat area, Φ = EA cosθ. Gauss's law: ∮E·dA = q_enclosed/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. If a closed surface encloses charge +2 microC, net flux through it is Φ = q/ε0 = (2 x 10^-6)/(8.85 x 10^-12) = 2.26 x 10^5 N m^2 C^-1 outward.

3-mark answer

For a small area element dA, the area vector is perpendicular to the surface. Electric flux is = 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. Electric flux: Φ = ∫E·dA, SI unit N m^2 C^-1. For uniform field over flat area, Φ = EA cosθ. Gauss's law: ∮E·dA = q_enclosed/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. A cube encloses a point charge of +8.85 nC. q = 8.85 x 10^-9 C. Net flux Φ = q/ε0 = (8.85 x 10^-9)/(8.85 x 10^-12) = 1.0 x 10^3 N m^2 C^-1. This is total flux through all six faces together. 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. 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.
Practice

Concept practice is coming soon

Join the waitlist for concept-level MCQs and weak-concept practice.

10 MCQs5 MinutesInstant Results
Join Waitlist for Practice

Help improve this page

Found something confusing, incorrect, or missing?