The flux through the hemispherical surface

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SUMMARY

The flux through a hemispherical surface in a uniform electric field is calculated using the formula πR²E, where R is the radius of the hemisphere and E is the electric field strength. The discussion clarifies that to apply Gauss's law, one must consider the closed surface formed by adding a flat disk to the hemisphere. The total flux through the closed surface equals zero, leading to the conclusion that the flux through the hemispherical surface is equal to the negative of the flux through the disk. This is a critical concept in electrostatics, particularly when dealing with non-closed surfaces.

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  • Understanding of Gauss's Law in electrostatics
  • Familiarity with electric flux and its mathematical representation
  • Knowledge of uniform electric fields and their properties
  • Basic geometry of hemispheres and disks
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  • Study the application of Gauss's Law for various geometries
  • Learn about electric field calculations for different charge distributions
  • Explore the concept of electric flux in non-closed surfaces
  • Investigate the relationship between area and surface area in three-dimensional shapes
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Students and professionals in physics, particularly those focusing on electromagnetism, as well as educators teaching concepts related to electric fields and flux calculations.

  • #31
gracy said:
What is surface area of disk?
A disk can be a three-dimensional body like a floppy disk http://www.merriam-webster.com/dictionary/disk. In that case, it is a flat right cylinder, having a side and two bases. The surface are is twice the area of the base plus the area of the side.
In your problem, the disk is a circle. You need to calculate the flux through the circle .
 
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  • #32
gracy said:
So,how surface area of sphere comes out to be 4 pi r^2?Does it include 4 circles?I don't think so
Gracy, I can not believe that you never learned the area of the sphere. It does not mean that it consists of 4 circles.
 
  • #33
gracy said:
So,how surface area of sphere comes out to be 4 pi r^2?Does it include 4 circles?I don't think so
It happens to have the same area as four circles of the same radius. But it's a curved surface, so you would have to cut it into infinitely many pieces to compose it of plane areas. Anyway, I was not suggesting the closed surface necessarily be cut into plane areas. You could cut it into eight octants, say, and each would turn out to have an area ##\pi r^2/2##. I was just trying to show you that there is no fundamental difference between surface area of a closed 3D region and area of a surface (which might or might not be a plane surface).
 
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  • #34
ehild said:
I can not believe that you never learned the area of the sphere.
Yes! I never heard ,I only knew surface area of sphere.
 
  • #35
gracy said:
Yes! I never heard ,I only knew surface area of sphere.
ehild is referring to the surface area of a sphere. There is an ambiguity here. The term sphere is used equally to mean a solid sphere or a spherical surface. The area of a spherical surface means the same as the surface area of a solid sphere. The qualifier "surface" area is only needed/appropriate when "sphere" means a solid sphere.
 
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  • #36
gracy said:
it should be πR^2E
I got why!Thanks haru!
 
  • #37
gracy said:
It was always there!
This does not help me at all. Sorry.
 
  • #38
BvU said:
This does not help me at all
What do you mean?
 
  • #39
BvU said:
This does not help me at all
Oh!understood.Even when you say/write something straight forward I tend to take it logically or sarcastically to be more precise.
 

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