Why is Flux Different for Spheres Compared to Disks in Gauss' Law?

[Brajesh kedia]

Flux through a closed surface is zero due to external charge but why it is not so in case of sphere

 

[Doc Al]

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]

But we know flux must be zero..for external charge

 

[Doc Al]

But we know flux must be zero..for external charge

Please explain what that diagram represents.

 

[Brajesh kedia]

Its a sphere of radius 3m and at distance 4m there is a charge q

 

[Doc Al]

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.

 

[Brajesh kedia]

Yeah but as i shown u the above solution flux is calculator to be q/E *(1-COS@)

 

[Brajesh kedia]

That round is sphere... and 4m far there is a q charge...and flux due to that charge is q/5Eo

 

[Doc Al]

Yeah but as i shown u the above solution flux is calculator 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.

 

[Doc Al]

Yeah but as i shown u the above solution flux is calculator 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 $(1 - \cos\theta)/2$.

 

[Doc Al]

Note that there's no "closed sphere" involved here.

 

[Brajesh kedia]

I did not understand ur points..i am trying to calculate the flux through the sphere by a point charge q

 

[Doc Al]

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.

 

[Brajesh kedia]

How to do that

 

[Dale]

Yeah but as i shown u the above solution flux is calculator 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.

 

[Doc Al]

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.

 

[Brajesh kedia]

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

 

[Nugatory]

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.

 

[Brajesh kedia]

Do u mean that the flux i calculated is the amt of flux entering the charge.Can u please explain how formula came

 

[Doc Al]

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.

Can u please explain how formula came

I have already done so up above. You may need to look up what a "steradian" is.

 

[Brajesh kedia]

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...

 

[Brajesh kedia]

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..?

 

[Doc Al]

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

 

[Doc Al]

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.

 

[Brajesh kedia]

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

 

[Brajesh kedia]

And thanks for link.well check it out soon when again working on with same lesson

 

[Doc Al]

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.

 

[rumborak]

Or put differently, a closed surface has no edge. An open one does.

 

[Brajesh kedia]

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..