Electric field of a part of a hemisphere

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SUMMARY

The discussion focuses on determining the electric field at point O from a section of a hemisphere using symmetry principles rather than Gauss's law. The key insight is that the section exhibits both longitudinal and transverse symmetry, which allows for the conclusion that the net electric field at O lies along the intersection of the two symmetry planes. Despite the presence of these symmetries, the application of Gauss's law is insufficient, prompting the use of an alternative argument proposed by Krotov to simplify the problem without integration.

PREREQUISITES
  • Understanding of electric fields and their calculations
  • Familiarity with Gauss's law and its applications
  • Knowledge of symmetry principles in physics
  • Basic concepts of hemispherical geometry
NEXT STEPS
  • Study the application of Gauss's law in various geometries
  • Research Krotov's method for simplifying electric field calculations
  • Explore longitudinal and transverse symmetry in electrostatics
  • Learn about electric field calculations in non-uniform charge distributions
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Physics students, educators, and professionals interested in electrostatics, particularly those studying electric fields in complex geometries and symmetry applications.

ermia
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Homework Statement
There is a problem in krotov which I need help solving it.(3.4)
the solution of krotov is not obvious for me.
The problem's statement is uploaded as an image.
It says we want to find the electric feild of a part of a hemisphere with angle alpha at its center.
Relevant Equations
Maybe nothing but vector summation.
I tried gauss law.
And the fact that if alpha is less than pi/2 we can say that we have two parts with angle alpha and one other part which has a normal field at the center.
But non of them helped me answer.

The problem's solution says that we can use the fact that our section has longitudinal and transverse symmetry. But how it has such a symmetry?
 

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1662929458596.png


Imagine a plane that is perpendicular to the diameter AB and also contains the line OC. By symmetry, the net electric field at O must lie in this plane. I think this might be the "transverse symmetry". The plane transversely cuts the region in half.

1662929615424.png


Imagine a plane that contains the diameter AB and bisects the angle ##\alpha##. This plane contains the colored region ADB. It also cuts the charged region in half. By symmetry, the net electric field at O must lie in this plane. I think this might be the "longitudinal" symmetry.

Anyway, whatever we call these planes, the net field at O must lie in both planes. That is, the field must lie along the intersection of these two planes.

Even though we have these two symmetry planes, there is not enough symmetry to use Gauss' law. But Krotov avoids integration by using a very clever argument.
 
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