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.
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Student-friendly explanation
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.
How to write this in exams
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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.
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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.
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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.
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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
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Examples and method
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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
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