# Electric field of a part of a hemisphere

• ermia
In summary, there is a problem involving Gauss' law and alpha being less than pi/2. The solution involves using the fact that there is longitudinal and transverse symmetry, which can be seen by imagining a plane perpendicular to the diameter AB and containing the line OC. Another plane can be imagined containing the diameter AB and bisecting the angle alpha, which also cuts the charged region in half. However, there is not enough symmetry to use Gauss' law, but Krotov uses a clever argument to avoid integration.
ermia
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?

#### Attachments

• Krotov.jpg
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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.

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.

Last edited:
ermia and topsquark

## 1. What is the electric field of a part of a hemisphere?

The electric field of a part of a hemisphere is a vector quantity that describes the strength and direction of the electric force experienced by a charged particle at any point within the region of the hemisphere.

## 2. How is the electric field of a part of a hemisphere calculated?

The electric field of a part of a hemisphere can be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the hemisphere, and r is the distance from the center of the hemisphere to the point where the electric field is being measured.

## 3. What factors affect the electric field of a part of a hemisphere?

The electric field of a part of a hemisphere is affected by the charge of the hemisphere, the distance from the center of the hemisphere, and the angle at which the electric field is being measured.

## 4. How does the electric field of a part of a hemisphere vary with distance?

The electric field of a part of a hemisphere follows an inverse square law, meaning that as the distance from the center of the hemisphere increases, the electric field decreases proportionally.

## 5. What is the direction of the electric field of a part of a hemisphere?

The direction of the electric field of a part of a hemisphere is always radial, meaning it points away from the center of the hemisphere. This direction can be determined using the right-hand rule, where the thumb points in the direction of the electric field and the fingers curl in the direction of the electric force on a positive test charge.

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