ElectromagnetismLesson 3 of 4 in the course Gauss's Law

Applying Gauss’s Law to Symmetric Charge Distributions

5 flashcards · 5 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

Gauss’s law relates electric flux through a closed surface to the charge enclosed by that surface. When a charge distribution has enough symmetry, this relationship can be used to determine the electric field.

Electric flux measures how much electric field passes through a surface. For a closed surface SS, Gauss’s law is

ΦE=∮SE⃗⋅n^ dA=qencε0.\Phi_E=\oint_S \vec{E}\cdot\hat{n}\,dA=\frac{q_{\mathrm{enc}}}{\varepsilon_0}.

All flashcards

What is electric flux?
Electric flux measures the electric field passing through a surface. For a small area element, it depends on the field component along the area’s normal.
State Gauss’s law.
The net electric flux through a closed surface equals the net charge enclosed divided by the permittivity of free space: ∮SE⃗⋅n^ dA=qenc/ε0\oint_S \vec{E}\cdot\hat{n}\,dA=q_{\mathrm{enc}}/\varepsilon_0.
What is a Gaussian surface?
An imaginary closed surface selected to use Gauss’s law. Its shape is chosen to match the charge distribution’s symmetry and simplify the flux calculation.
Which three symmetries are especially useful for applying Gauss’s law?
Spherical, cylindrical, and planar symmetry.
Why must enclosed charge be determined carefully?
Gauss’s law uses only the net charge inside the selected closed surface. Charges outside the surface do not contribute to qencq_{\mathrm{enc}}.

Quiz questions

  1. For a small area element on a closed surface, when is its electric-flux contribution positive?Answer: When the electric field points along the outward normal
  2. Outside a spherically symmetric charge distribution, what determines the electric field?Answer: The total charge, as if concentrated at the center
  3. How does the electric-field magnitude inside a uniformly charged solid sphere vary with distance from its center?Answer: It increases linearly with distance
  4. Why is the electric flux through the end caps of a coaxial cylindrical Gaussian surface zero for a cylindrically symmetric charge distribution?Answer: The radial field is parallel to the end-cap surfaces
  5. How does the field magnitude of an ideal infinite uniformly charged plane vary with perpendicular distance from the plane?Answer: It is independent of that distance

Based on University Physics Volume 2 by Samuel J. Ling, William Moebs, Jeff Sanny (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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