Is Gauss' Law Summation Necessary for Uniform Electric Fields?

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SUMMARY

The discussion centers on the application of Gauss' Law in calculating electric flux through a circular surface in a uniform electric field. The problem involves a circular surface with a radius of 0.072m exposed to an electric field of 1.44x104 N/C, resulting in an electric flux of 82 Nm2C-1. The angle between the electric field direction and the surface normal is calculated to be 69.5 degrees. The key takeaway is that in uniform electric fields, both the electric field strength (E) and the cosine of the angle (cos φ) are constant, allowing them to be factored out of the summation.

PREREQUISITES
  • Understanding of Gauss' Law and electric flux
  • Familiarity with the concept of uniform electric fields
  • Knowledge of trigonometric functions, specifically cosine
  • Ability to perform basic calculus operations, such as integration over area
NEXT STEPS
  • Study the implications of Gauss' Law in non-uniform electric fields
  • Learn about electric field lines and their relation to electric flux
  • Explore the derivation of electric flux equations in various geometries
  • Investigate applications of Gauss' Law in real-world scenarios, such as capacitors
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone seeking to deepen their understanding of electrostatics, particularly in the context of Gauss' Law and electric fields.

BOAS
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Hello,

i'm doing some practice problems using Gauss' law, but I feel like my work is 'sloppy'. I'll show an example, where I think I get the right answer, but it feels like I'm neglecting to treat the summation properly, or perhaps I don;t quite understand why what I'm doing is fine...

Homework Statement



A circular surface with a radius of [itex]0.072m[/itex] is exposed to a uniform external electric field of magnitude [itex]1.44[/itex]x[itex]10^{4}NC^{-1}[/itex]. The electric flux through the surface is [itex]82 Nm^{2}C^{-1}[/itex]. What is the angle between the direction of the electric field and the normal to the surface?

Homework Equations





The Attempt at a Solution



[itex]\Phi_{E} = \Sigma(E \cos \phi) \Delta A[/itex]

[itex]\Sigma \Delta A = \pi r^{2}[/itex]

[itex]\cos \phi = \frac{\Phi_{E}}{E \pi r^{2}} = 0.3497[/itex]

[itex]\phi = \cos^{-1}(0.3497) = 69.5 \deg[/itex]

feel like I've done this correctly, but I also feel like I've just dropped the summation sign without really knowing why. I know I summed up all the little areas to give the area of a circle, but it was also the sum of [itex]E \cos \phi[/itex].

Could some please explain why I don't actually do a summation there?

This may be a bit of a bizarre question...

Thanks!
 
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BOAS said:
Hello,

i'm doing some practice problems using Gauss' law, but I feel like my work is 'sloppy'. I'll show an example, where I think I get the right answer, but it feels like I'm neglecting to treat the summation properly, or perhaps I don;t quite understand why what I'm doing is fine...

Homework Statement



A circular surface with a radius of [itex]0.072m[/itex] is exposed to a uniform external electric field of magnitude [itex]1.44[/itex]x[itex]10^{4}NC^{-1}[/itex]. The electric flux through the surface is [itex]82 Nm^{2}C^{-1}[/itex]. What is the angle between the direction of the electric field and the normal to the surface?

Homework Equations





The Attempt at a Solution



[itex]\Phi_{E} = \Sigma(E \cos \phi) \Delta A[/itex]

[itex]\Sigma \Delta A = \pi r^{2}[/itex]

[itex]\cos \phi = \frac{\Phi_{E}}{E \pi r^{2}} = 0.3497[/itex]

[itex]\phi = \cos^{-1}(0.3497) = 69.5 \deg[/itex]

feel like I've done this correctly, but I also feel like I've just dropped the summation sign without really knowing why. I know I summed up all the little areas to give the area of a circle, but it was also the sum of [itex]E \cos \phi[/itex].

Could some please explain why I don't actually do a summation there?

This may be a bit of a bizarre question...

Thanks!
You've done it correctly. [itex]E[/itex] and [itex]\cos \phi[/itex] are constant over the area because the electric field is uniform, so they can come out of the summation.

Chet
 
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