How to Calculate Electric Field of Half Spherical Shell with Gauss Law?

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SUMMARY

The discussion focuses on calculating the electric field of a half spherical shell with radius R and uniform surface charge density σ using Gauss's Law. Participants suggest that while the Gauss surface cannot be closed above the half spherical shell, one can derive the electric field by first calculating it for a full spherical shell and then halving the result. The analogy of a salad bowl is used to visualize the charge distribution along the axis of symmetry.

PREREQUISITES
  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric field concepts and calculations
  • Knowledge of spherical symmetry in charge distributions
  • Basic integration techniques for electric field calculations
NEXT STEPS
  • Study the application of Gauss's Law to different charge distributions
  • Learn how to calculate electric fields for full spherical shells
  • Explore the concept of electric field lines and their implications
  • Investigate the use of numerical methods for complex charge configurations
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in understanding electric fields generated by charged surfaces.

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Homework Statement



Hey guys.
I got this question:
"Calculate the electric field of half a spherical shell with radius R and uniform surface charge density \sigma along its axis of symmetry above the shell."
I guess I shell use Gauss law, the thing is that the Gauss surface will not be closed above the half spherical shell so I though about calculating the electric field for a full spherical shell and then divide it by 2 to get the electric field above the upper half spherical shell.
What do you say? is that the way to go?

Thanks in advance.

Homework Equations





The Attempt at a Solution


 
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If you're talking about losing the half away from the side that is charged, I'd think not.

From the statement I take it that the points along the axis make the charges look like concentric rings spaced at a distance according to their spherical geometry away from whatever point you are at. Like looking down at a salad bowl turned face down?
 
LowlyPion said:
If you're talking about losing the half away from the side that is charged, I'd think not.

From the statement I take it that the points along the axis make the charges look like concentric rings spaced at a distance according to their spherical geometry away from whatever point you are at. Like looking down at a salad bowl turned face down?

Yeah, it's like a salad bowl turned face down :smile:
I need to find a closed Gauss surface that will contain this "bowl" and I can't think of one.

I'm sorry, I didn't understand what you mean by "If you're talking about losing the half away from the side that is charged, I'd think not.".

Thanks a lot.
 
I didn't understand what you mean by ...
I was simply referring to your notion that you would take the integral of a whole sphere and then take away the mirror of the salad bowl.
 

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