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  5. 05Revise your understanding
  6. 06Repeat with spacing
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Gauss's Law · Lesson 2 of 5
LESSON
Flux through a closed surface

Electric flux measures the number of field lines passing through a surface. For a closed surface, Gauss's law states that the total flux is proportional to the charge enclosed:

ΦE=EdA=Qencε0\Phi_E = \oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}

Why it's powerful. Whenever you can exploit symmetry — spherical, cylindrical, planar — Gauss's law turns hard integrals into simple algebra.

A spherical Gaussian surface surrounding a point charge, showing radius R, an area element dA, and the outward electric field
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Assistant

Stuck on something? Ask mid-lesson. It explains concepts the way your textbook does — and links the exact lesson it's drawn from.

Flux through a closed surface
University Physics
Why is the flux the same for any closed surface around the charge?

Because flux counts the field lines passing through the surface, and every line from the charge inside crosses any closed surface around it exactly once. So the total depends only on the enclosed charge:

ΦE=Qencε0\Phi_E = \dfrac{Q_{\text{enc}}}{\varepsilon_0}

The surface's size and shape don't change it.

Flux through a closed surface
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Flashcards

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Key terms · 4 cards
FLASHCARDSKey terms · 4 cards
FRONT
Electric flux
BACK
The number of electric field lines passing through a surface.ΦE=EdA\Phi_E = \int \vec{E}\cdot d\vec{A}
Tap to flip · ← → to navigate

Practice what you learned

After each lesson, practice the core concepts to make them stick.

Gauss's Law · Q2 of 4
FILL IN THE BLANKQ2 of 4
Gauss's law states that the electric flux through a closed surface equals the total enclosed, divided by ε0\varepsilon_0.
charge
field
potential
current
Check

Challenge Questions

Solve an open-ended problem by hand, AI grades your reasoning. Challenges are skippable — the ones you skip wait for you at the end of the chapter, so you're never blocked mid-lesson.

Gauss's Law · Q4 of 4Backlog · 2
CHALLENGE

A point charge +q sits at the center of a spherical Gaussian surface of radius R. Use Gauss's law to derive the magnitude of the electric field at the surface, and explain why your result is independent of R.

Solution_q4.jpg
Hand-written work · 1.2 MB
Graded — correct

Excellent. You correctly applied symmetry to take E outside the integral, used the surface area 4πR² for the sphere, and noted that R cancels in the final expression.

Skip · do it at the end of the chapter
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Look up terms

Look up terms and definitions without interrupting your work.

Definition lookup
DEFINITION LOOKUP
…choose a Gaussian surface that exploits the symmetry of the charge distribution to make EdA\oint \vec{E}\cdot d\vec{A} tractable.
Gaussian surface
Gaussian surface
From Ch 6 · Gauss's Law · Section 1
An imaginary closed surface chosen so that the electric field has the same magnitude over the entire surface, or is perpendicular or parallel to it. This symmetry lets you pull E outside the flux integral and solve for it directly.

Spaced Repetition

Every question you answer gets scheduled for review.

Review · Electricity & Magnetism
REVIEW12 due today
Ch 2 · Gauss's Law
Electric flux through a closed surface is proportional to the enclosed.
Correct · Next review · in 3 days
1d3d9d3w
Spacing grows each time you get it right.

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