This lesson uses a point charge to show how electric flux behaves through a closed spherical surface. The result reveals an important relationship between the strength of the electric field and the size of the surface.
E=4πε01r2qr^
Consider a spherical surface of radius R centered on the charge. At every point on this sphere, the outward unit normal n^ points in the same direction as r^, and the field has the same magnitude. The flux through a small area dA is therefore
All flashcards
What is the electric flux through a sphere centered on a point charge q?
It is 4q/ε0.
How does a point charge’s electric field magnitude vary with distance?
It is proportional to 1/r2.
Why is flux through a sphere centered on a point charge independent of the sphere’s radius?
The field varies as 1/R2, while the sphere’s area varies as R2; these factors cancel.
How do field lines illustrate constant flux through concentric spheres?
The lines cross both spheres but spread over more area on the larger sphere, lowering their density.
Does zero net flux mean the electric field is zero everywhere on a surface?
No. Inward and outward flux contributions can cancel even when the field is nonzero on the surface.
Quiz questions
Which equation states Gauss’s law for a closed surface S?Answer: ∮SE⋅n^dA=qenc/ε0
Which electric field belongs in the surface integral of Gauss’s law?Answer: The total electric field at each surface element, including contributions from inside and outside charges.
What determines the enclosed charge qenc for a Gaussian surface?Answer: The algebraic sum of charges inside the surface.
What is the sign of the net flux through a closed surface enclosing a net negative charge?Answer: Negative, using the outward normal convention.
What shape must a Gaussian surface have for Gauss’s law to apply?Answer: Any closed shape; it need not be spherical.
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.