Can Gauss' Law be applied to point charges on non-spherical surfaces?

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SUMMARY

Gauss' Law can be applied to point charges located near non-spherical surfaces, such as cubes and squares. In the discussion, a point charge of 1.84 microC is placed at the center of a cubical Gaussian surface, and the electric flux (\Phi_E) through the surface is calculated using the formula \(\oint \vec{E} d\vec{A} = \frac{q_{inc}}{\epsilon_0}\). The participants also explore calculating flux through a square surface when a point charge is positioned directly above its center. The conclusion emphasizes that while the shape of the surface does not affect the total flux, understanding the implications of Gauss' Law is crucial for accurate calculations.

PREREQUISITES
  • Understanding of Gauss' Law and its mathematical formulation
  • Familiarity with electric flux concepts
  • Basic knowledge of point charges and their properties
  • Ability to perform calculations involving microcoulombs and electric fields
NEXT STEPS
  • Study the application of Gauss' Law to various geometries, including cylindrical and spherical surfaces
  • Learn about electric field calculations for point charges using integration techniques
  • Explore the concept of electric flux in non-uniform electric fields
  • Investigate the implications of symmetry in electric field calculations
USEFUL FOR

Students of electromagnetism, physics educators, and anyone interested in applying Gauss' Law to complex geometrical configurations in electrostatics.

manenbu
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2 questions:

1. A point charge of 1.84 microC is at the center of a cubical Gaussian surface 55cm on edge. Find \Phi_E through the surface.

So here I was thinking, well the shape doesn't matter so the surface can be a sphere, so I calculated it for a sphere and it was correct (taking the radius as half of the edge).
But - how to do it for a square?

2. This one is related too - being "A point charge +q is a distance d/2 from a square surface of side d and is directly above the center of the square. Find the flux through the square".

More or less the same, solve with letters instead of numbers and divide by six. Again - how to do it without turning it into a sphere?
 
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I think the best way to understand this is to think about what gauss' law says.

The equation is (of course):
<br /> \oint \vec{E} d\vec{A} = \frac{q_{inc}}{\epsilon_0}<br />
But what does this mean?
 
It means that the flux is equal to the charge inside the surface over e0.
Ok. It came to me now. :)
 

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