Electric Flux and Gauss's Law
Electric flux measures the signed electric field crossing a surface. Gauss's law relates net outward flux through a closed surface to the net enclosed charge. A surface chosen to match the charge distribution's symmetry can help calculate the field.
What a learner can do afterwards
- Calculates the flux through a flat surface for a uniform field at any angle
- States Gauss's law and explains why an outside charge contributes zero net flux through a closed surface
- Chooses a closed surface whose symmetry matches a given charge distribution
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Electric flux Φ measures the signed electric field crossing a surface. For a flat patch in a uniform field, Φ = EA cosθ, in N·m²/C. E is field strength; A is area. θ is the angle between the field and a chosen normal, perpendicular to the patch. E cosθ is the signed field component along that normal. Use a calculator in degree mode to evaluate cosθ. cos0° = 1, cos90° = 0 and cos180° = -1. A field parallel to the patch gives zero flux.
A uniform field has strength 30 N/C. A flat patch has area 0.4 m², and its chosen normal makes an angle of 60° with the field. Since cos60° = 0.5, its flux is 30 × 0.4 × 0.5 = 6 N·m²/C. Use the angle to the normal, not the angle to the patch itself.
For a curved surface, add E cosθ dA over tiny patches; the limit is a surface integral. On a closed surface, normals point outward: leaving flux is positive and entering flux is negative. Gauss's law is Φ = Qenc/ε0. Qenc is the signed sum of enclosed charges; ε0 is vacuum permittivity. This gives net outward flux for any closed shape. Avoid point charges on the boundary. A continuous line may cross the surface with zero finite charge at its crossing points.
An outside charge can change the field at every patch, but its contribution to the total closed-surface flux is zero. In a field-line picture, its inward and outward crossings cancel. A charge-free box in a uniform field therefore has zero net flux, even though its field need not be zero. Positive net flux means net positive enclosed charge, not that every enclosed charge is positive.
Center a sphere of radius r on an isolated positive point charge Q. The field is normal to the sphere and has the same strength everywhere on it. The area is 4πr² and E = Q/(4πε0r²), so Φ = E(4πr²) = Q/ε0. Doubling r multiplies area by 4 and field strength by 1/4. The flux stays fixed.
Use a surface matched to the charge distribution's symmetry. An isolated point charge uses a centered sphere. A uniform infinite line uses a closed coaxial cylinder. Its field has constant strength and is normal to the curved side; end-cap flux is zero. Side flux is EA cosθ: θ = 0° for a positive line and 180° for a negative line, so flux is +EA or -EA. A is the side area. Gauss's law also holds when symmetry cannot make E constant.
Use Φ = EA cosθ for a uniform field and flat patch. Closed-surface net flux is Qenc/ε0. Symmetry can turn the flux integral into an equation for E.
Lesson, assessment questions and revised flashcards written for this dataset. Physics facts and teaching structures were studied in OpenStax University Physics Volume 2, sections 6.1, 6.2 and 5.6.
Readings and revised flashcards written for this dataset. Physics facts and teaching structures were studied in OpenStax University Physics Volume 2, sections 6.1, 6.2 and 5.6. Video: Conceptual understanding of flux, credited to Khan Academy.
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