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Physics · University

Electric flux and Gauss symmetry

Electric flux is the oriented surface integral of E·dA. A negative flux reflects direction relative to the chosen normal, not negative area.

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Prerequisites

Vector fields, dot products and surface integrals.

Study notes

Flux

Electric flux is the oriented surface integral of E·dA. A negative flux reflects direction relative to the chosen normal, not negative area.

Gauss law

For a closed surface, total electric flux equals enclosed charge divided by ε0. Charges outside can affect the local field but not net enclosed charge.

Symmetry

Gauss law always holds classically, but extracting E from it easily requires useful symmetry. Do not assume E is constant on an arbitrary Gaussian surface.

Conductors

In electrostatic equilibrium the electric field inside an ideal conductor material is zero. Cavities require separate reasoning about enclosed charges.

Worked examples

A uniform field of 4 N/C is parallel to a planar area's outward normal. For area 3 m², find flux.

Flux EA cos(0)=12.

12 N m²/C

What charge enters the right side of Gauss law for a closed surface?

The enclosed algebraic charge controls total flux.

Net enclosed charge

Check your understanding

  1. Use ε0=2 in stated model units. A closed surface encloses charge 10. Find total electric flux in the matching units.

  2. Does zero net flux imply zero electric field at every point on the surface?

    1. Yes, in every geometry
    2. Only if charge exists outside
    3. Yes, if the area is nonzero
    4. No; contributions can cancel

  3. Explain why a point charge outside a Gaussian surface can affect E without changing its total flux.

  4. A field makes an angle of 60 degrees with the normal. If EA=10 N m²/C, find flux.

Answers and marking guidance

Answers and marking notes
  1. 5

    Gauss law gives Qenclosed/ε0=10/2; the units are deliberately rescaled.

  2. No; contributions can cancel

    A uniform external field can enter and leave a closed surface with zero net flux.

  3. It contributes a local field, but its field lines enter and leave the closed surface with cancelling oriented contributions. It adds no enclosed charge.

    Credit the distinction between field and integral flux.

  4. 5 N m²/C

    Multiply EA by cos 60°=1/2.

Further reading and review references

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