Electric field of a part of a hemisphere

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The discussion revolves around applying Gauss's law to determine the electric field at point O within a hemisphere section. It highlights the significance of longitudinal and transverse symmetry in analyzing the electric field, emphasizing that the net electric field must lie in the intersection of two specific symmetry planes. Despite the presence of these symmetry planes, the discussion concludes that Gauss's law cannot be directly applied due to insufficient symmetry. Instead, a clever argument by Krotov is mentioned as a means to avoid integration for solving the problem. The conversation underscores the complexities involved in calculating electric fields in geometrically symmetric configurations.
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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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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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