Gauss Law for non uniform spherical shell

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  • #1
nabeel17
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So i can see by symmetry arguments why The electric field inside a uniformly charged spherical shell would be zero inside.

But what about a non uniformly charged spherical shell. Say most of the charge is located on one side, why is the electric field still zero? I can see that the flux through any Gaussian surface I draw inside may be zero but when I look at it I don't really understand why the electric field is 0. If most the charge is located on side, I would imagine that a test charge inside would either attract or be repelled since the charge is not uniform. Can someone clear this up for me?
 

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Simon Bridge
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Well - consider the spherical shell is non-uniformly charged as follows ... all the charge is concentrated in a small spot on one side, and the rest is neutral. What is the field?

I think you need to ask - is this a conducting shell, or an insulator?
 
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D H
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Gauss's law does not say that the electrical field inside a non-uniform spherical shell is zero. Consider an extreme case of a spherical shell that contains a point charge, with the rest of the shell vacuum. Is the electrical field inside that shell zero? Of course not.
 
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Redbelly98
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So i can see by symmetry arguments why The electric field inside a uniformly charged spherical shell would be zero inside.
Symmetry arguments say the field is zero at the center. To say it's zero everywhere inside requires evaluating an integral, which turns out to give zero everywhere inside. This is the Shell Theorem.
 
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nabeel17
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Ok I can see how the electric field is 0 inside if the sphere is uniformly charged but what if the sphere is not. What if say most of the charge is in one hemisphere?
 
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D H
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A charged hemisphere looks like a point charge from very far away. Close up, you'll need to use spherical harmonics.
 
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Vanadium 50
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I think you need to ask - is this a conducting shell, or an insulator?

This was important.
 
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nabeel17
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Gauss's law does not say that the electrical field inside a non-uniform spherical shell is zero. Consider an extreme case of a spherical shell that contains a point charge, with the rest of the shell vacuum. Is the electrical field inside that shell zero? Of course not.


Right but doesnt Gauss' law say that the flux (EA) = charge enclosed. So say i have a nonuniformly charged shell (maybe not the extreme case but charged more on side then another), the charge enclosed is zero so the electric field is zero? Or am I wrong about that...?
 
  • #9
nabeel17
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Well - consider the spherical shell is non-uniformly charged as follows ... all the charge is concentrated in a small spot on one side, and the rest is neutral. What is the field?

I think you need to ask - is this a conducting shell, or an insulator?

Here i can treat it as a point charge? But then the electric field inside is not zero? Talking about an insulator. The flux equals the charge enclose/epsilon and if the charge enclosed is zero, the E field should be zero inside
 
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Doc Al
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Right but doesnt Gauss' law say that the flux (EA) = charge enclosed. So say i have a nonuniformly charged shell (maybe not the extreme case but charged more on side then another), the charge enclosed is zero so the electric field is zero? Or am I wrong about that...?
Yes, you are wrong about that. Just because the flux is zero does not mean that the field at any particular point on the surface is zero. The flux is the integral of E*dA over the whole surface; that total will be zero, but the field can certainly be nonzero at various points.
 
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  • #11
Simon Bridge
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Right but doesnt Gauss' law say that the flux (EA) = charge enclosed.
No it doesn't.
I've been caught out like that too ;)

If you have a point charge (never mind the sphere) you know the field is non-zero everywhere right? Yet it is trivial to draw a surface that does not enclose the charge. Does that mean that suddenly the field is zero everywhere? This is a contradiction so something is wrong. Gauss was smart enough to spot something like that so...

Gausses law actually says that the net flux through a Gaussian surface is proportional to the charge enclosed.

If there are no charges enclosed, then all the flux that enters the region must leave it at some point. If flux goes in one side and out the other, the net flux is zero but there is still flux in the enclosed region and, therefore, a non-zero electric field inside.
 
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