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Brajesh kedia
Flux through a closed surface is zero due to external charge but why it is not so in case of sphere
Why do you think a sphere is treated differently than any other closed surface?Brajesh kedia said:why it is not so in case of sphere
Please explain what that diagram represents.Brajesh kedia said:But we know flux must be zero..for external charge
I don't see any sphere in the diagram. But what about it?Brajesh kedia said:Its a sphere of radius 3m and at distance 4m there is a charge q
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.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##.Brajesh kedia said:Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)
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.Brajesh kedia said:I did not understand ur points..i am trying to calculate the flux through the sphere by a point charge q
Please show your work step by step. Looking at the diagram I have no idea how you came to this conclusion.Brajesh kedia said:Yeah but as i shown u the above solution flux is calculater to be q/E *(1-COS@)
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.Brajesh kedia said:How to do that
Brajesh kedia said:but some are saying no
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:Do u mean that the flux i calculated is the amt of flux entering the charge.
I have already done so up above. You may need to look up what a "steradian" is.Brajesh kedia said:Can u please explain how formula came
You need to understand solid angles and steradians. Start here: SteradianBrajesh 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...
Gauss' law applies to a closed surface, not a disk. The flux through a disk would not be zero.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..?
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.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
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.
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.
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).
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.
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.