Exploring Gauss' Law: Understanding Spheres and Flux in Closed Surfaces

In summary, a closed surface is one that completely encloses a volume, while an open surface does not. A disc is an open surface that does not enclose a volume, while a sphere is a closed surface that encloses a volume. The flux through a closed surface can be calculated using Gauss' law, while the flux through an open surface cannot.
  • #1
Brajesh kedia
Flux through a closed surface is zero due to external charge but why it is not so in case of sphere
 
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  • #2
Brajesh kedia said:
why it is not so in case of sphere
Why do you think a sphere is treated differently than any other closed surface?
 
  • #3
But we know flux must be zero..for external charge
 

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  • #4
Brajesh kedia said:
But we know flux must be zero..for external charge
Please explain what that diagram represents.
 
  • #5
Its a sphere of radius 3m and at distance 4m there is a charge q
 
  • #6
Brajesh kedia said:
Its a sphere of radius 3m and at distance 4m there is a charge q
I don't see any sphere in the diagram. But what about it?

If there is a sphere--or any other closed shape--and there is a charge outside the sphere, then the total flux through the surface of that sphere will be zero.
 
  • #7
Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)
 
  • #8
That round is sphere... and 4m far there is a q charge...and flux due to that charge is q/5Eo
 
  • #9
Brajesh kedia said:
Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)
I think what you're trying to calculate is the amount of flux from the charge through the solid angle subtended by a cone of angle θ. That's not the net flux through a closed surface.
 
  • #10
Brajesh kedia said:
Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)
The solid angle would be ##2\pi (1 - \cos\theta)## steradians. The total solid angle for a sphere is ##4\pi## steradians. So the fraction of the total flux from that charge going into that solid angle would be ##(1 - \cos\theta)/2##.
 
  • #11
Note that there's no "closed sphere" involved here.
 
  • #12
I did not understand ur points..i am trying to calculate the flux through the sphere by a point charge q
 
  • #13
Brajesh kedia said:
I did not understand ur points..i am trying to calculate the flux through the sphere by a point charge q
Ah, I think I see what you are saying. That thing which is subtending the cone of flux (with half angle ##\theta##) is a sphere. To apply Gauss' law to that sphere you must calculate the total flux through the spherical surface. What you've attempted to calculate so far is just the flux into the sphere (through the side facing the charge). You must also consider the flux going out of the sphere (through the other side of the sphere). The total flux through that spherical surface will be zero.
 
  • #14
How to do that
 
  • #15
Brajesh kedia said:
Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)
Please show your work step by step. Looking at the diagram I have no idea how you came to this conclusion.
 
  • #16
Brajesh kedia said:
How to do that
If you draw yourself a proper diagram I think you'll see that no calculation is needed to show that the net flux through the sphere is zero.

Do this. Put the charge at the origin (0,0). Then draw a sphere (a circle on your diagram) centered somewhere along the x-axis. Then draw the tangents to the circle that pass through the origin. Those will define a cone of flux from the charge at the origin. You can calculate the angle of the cone. Where the cone intersects the sphere splits the sphere into two (unequal) pieces. Clearly, whatever flux enters one side of the sphere must exit the other, so the net flux is zero.

If you wish to answer a different question, such as "how much flux from the charge hits the sphere?", then you would proceed as I outlined above. I believe that's what you were attempting to calculate yourself.
 
  • #17
Yeah i believe net flux is zero but some are saying no..i am confused...even i knew about q=1/E not the formula which i was asked to use..the one i used
 
  • #18
Brajesh kedia said:
but some are saying no

Who is saying "no" and where? Everyone in this thread seems to be saying (with remarkable patience, I might add) that the flux through the surface of a sphere when there is no charge inside of it is zero.
 
  • #19
Do u mean that the flux i calculated is the amt of flux entering the charge.Can u please explain how formula came
 
  • #20
Brajesh kedia said:
Do u mean that the flux i calculated is the amt of flux entering the charge.
I believe, as I already had stated, that you were attempting to calculate the amount of flux from the point charge hitting (entering) the sphere.

Brajesh kedia said:
Can u please explain how formula came
I have already done so up above. You may need to look up what a "steradian" is.
 
  • #21
When i checked for what solid angle is i found it to be just 2pie(1-cos@) which is quite unsatisfactory and got an explanation that its a 3-D angle made by the area of cone,why,how,reason did not receive answer to such question...
 
  • #22
I would like to confirm that flux through a disc is also zero if the figure i drawn is disc (say just for now actually its a sphere for my questions) as some persons confused me that for disc it would not be zero..?
 
  • #23
Brajesh kedia said:
When i checked for what solid angle is i found it to be just 2pie(1-cos@) which is quite unsatisfactory and got an explanation that its a 3-D angle made by the area of cone,why,how,reason did not receive answer to such question...
You need to understand solid angles and steradians. Start here: Steradian
 
  • #24
Brajesh kedia said:
I would like to confirm that flux through a disc is also zero if the figure i drawn is disc (say just for now actually its a sphere for my questions) as some persons confused me that for disc it would not be zero..?
Gauss' law applies to a closed surface, not a disk. The flux through a disk would not be zero.
 
  • #25
I would be glad if u would make it easier for me to understand a difference between a closed surface and open surface and how disc and sphere are said so
 
  • #26
And thanks for link.well check it out soon when again working on with same lesson
 
  • #27
Brajesh kedia said:
I would be glad if u would make it easier for me to understand a difference between a closed surface and open surface and how disc and sphere are said so
Think of a closed surface as enclosing a volume and having an 'inside' and an 'outside'. A sphere encloses a volume, but a disk does not.
 
  • #28
Or put differently, a closed surface has no edge. An open one does.
 
  • #29
Disk has edge...i did not get that??
And does not the disc have inside and outside as if i say that charges enter the disc straight into center(say) and then straight outside then the amount of flux entered went outside..
 

1. What is Gauss' Law and why is it important?

Gauss' Law is a fundamental principle in the study of electromagnetism that describes the relationship between electric charges and electric fields. It states that the electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space. This law is important because it allows us to calculate electric fields in situations where symmetry is present, making it a powerful tool for solving complex problems.

2. How does Gauss' Law apply to spheres?

Gauss' Law can be applied to any closed surface, including spheres. In the case of a sphere, the electric flux through the surface is equal to the charge enclosed by the sphere divided by the permittivity of free space. This allows us to calculate the electric field at any point outside or inside the sphere, as long as the charge distribution is symmetric.

3. What is electric flux and how is it related to Gauss' Law?

Electric flux is a measure of the amount of electric field passing through a given surface. It is calculated by taking the dot product of the electric field and the surface area vector. In the context of Gauss' Law, it is related to the enclosed charge and permittivity of free space through the equation: electric flux = (enclosed charge) / (permittivity of free space).

4. Can Gauss' Law be used for non-symmetric charge distributions?

No, Gauss' Law can only be used for charge distributions that exhibit some form of symmetry. This is because the law relies on the assumption that the electric field is constant over the surface of the closed surface being considered. If there is no symmetry, the electric field will vary and the law cannot be applied.

5. What are some real-world applications of Gauss' Law?

Gauss' Law has numerous real-world applications, including in the design of electronic devices, such as capacitors and transistors. It is also used in the analysis of electric fields in conductors, semiconductors, and dielectrics. Additionally, it is a crucial component of Maxwell's equations, which describe the behavior of electromagnetic waves and are essential in fields such as telecommunications and electrical engineering.

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