How Well Do You Understand Gauss's Law?

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SUMMARY

This discussion centers on Gauss's Law, represented mathematically as \oint \vec{E}\cdot d\vec{a} =\frac{Q_{enc}}{\epsilon_0}. The participants analyze various statements regarding the implications of this law, ultimately concluding that option C is correct, as the integral equals zero when the enclosed charge is zero. They also dismiss options B, D, and E based on logical reasoning about electric fields created by external charges. The confusion around option A highlights the broader definition of the electric field, which includes contributions from all charges, not just those enclosed by the Gaussian surface.

PREREQUISITES
  • Understanding of Gauss's Law and its mathematical formulation
  • Familiarity with electric fields and their properties
  • Knowledge of electric charge distributions, including point charges and dipoles
  • Basic calculus for interpreting integrals in physics
NEXT STEPS
  • Study the implications of electric dipoles on electric fields
  • Learn about the applications of Gauss's Law in various geometries
  • Explore the concept of electric field lines and their relationship to charge distributions
  • Investigate the role of external charges in influencing electric fields within Gaussian surfaces
USEFUL FOR

Students of electromagnetism, physics educators, and anyone seeking a deeper understanding of electric fields and Gauss's Law applications.

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Consider Gauss's law: [tex] \oint \vec{E}\cdot d\vec{a} =\frac{Q_{enc}}{\epsilon_0}[/tex] . Which of the following is true?

A. E Must be the electric field due to the enclosed charge

B. If q=0 then E=0 everywhere on the Gaussian surface

C. If the charge inside consists of an electric dipole, then the integral is zero

D. E is everywhere parallel to dA along the surface

E. If a charge is placed outside the surface, then it cannot affect E on the surface.
The attempt at a solution

I've ruled out B: Because I can have a point charge outside the Gaussian surface and so E is not zero necessarily at the surface since it will create an E field.

I've ruled out D: Because I can have a cube and E will not always be parallel to the 6 sides. Only case I can think of E being always parallel to dA is for a sphere.

I've ruled out E: Because this is similar to B. The external charge will create an E field of E=kQ/r^2.

So I say the answer is C since q(enclosed) will be zero leaving the integral equal to zero. Or, the answer could be A since isn't that kind of the definition of Gauss's law anyways? Or well, I guess not since we could have a Gaussian surface with no charge in it and a charge outside with E=kQ/r^2. So, the E vector in the integral is not necessarily due to the charge inside the Gaussian surface, right?

I don't think I'm supposed to have multiple answers though...not sure. I'm leaning more towards answer C.
 
Last edited:
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I don't like A. E is the electric field due to all charges in the universe, not just the ones inside. As you said in your comment on B.
 

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