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Conducting Shell

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In a conducting shell,with inner radius R1 and outer radius R2,and with charge Q at the centre,the Potential at surface is (kQ/R2),Why it is not (KQ/R1)?? :confused:
 

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Meir Achuz
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All the charge on a conducting shell is on the outer surface.
 
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Meir,
the charge Q present at the centre will induce -Q at the inner surface of shell and +Q at the outer surface of shell.
So obviously there is charge on inner surface too.!
 
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There are charge on inner surface but is canceled by the center charge, and there are charge Q distribute on the surface. So we can see the shell as a charged shell that charge Q is on the surface.
 
Meir Achuz
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brasil is right. I misread the original question, and didn't realize there was a charge at r=0.
 
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That's okay Brasil but howcome,the same potential for outer surface becomes equal to that of inner surface.!
 
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We can see potential as work done by electric field (albeit this "work" isn't the same as Force do) There's no field inside the shell, hence the potential is all the same from the outer surface to the center.
 
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brasilr9 said:
We can see potential as work done by electric field (albeit this "work" isn't the same as Force do) There's no field inside the shell, hence the potential is all the same from the outer surface to the center.
correctamundo

ps : do you guys know the charge distribution in the shell if the inner charge is not at the center of the sphere (let us say it is 1 cm away from the center to the left side)...what is the charge distribution in the shell ? is it uniform ? this is a classic... :tongue2:

marlon
 
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Of course it isn't uniform
 
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brasilr9 said:
Of course it isn't uniform
WRONG

marlon
 
SpaceTiger
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brasilr9 said:
We can see potential as work done by electric field (albeit this "work" isn't the same as Force do) There's no field inside the shell, hence the potential is all the same from the outer surface to the center.
Uh, if I'm visualizing this correctly, it's the same from the outer surface to the inner surface (since you're inside a conductor), but inside the inner surface, the potential goes as 1/r.
 
SpaceTiger
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marlon said:
ps : do you guys know the charge distribution in the shell if the inner charge is not at the center of the sphere (let us say it is 1 cm away from the center to the left side)...what is the charge distribution in the shell ? is it uniform ? this is a classic...
The charge distribution on the inner surface would have to be non-uniform in order for there to be an electric field of zero inside the conducting shell. However, the distribution on the outer surface would be uniform. Conductors essentially "hide" the information about the charge inside them and that's why the http://www.absoluteastronomy.com/encyclopedia/f/fa/faraday_cage.htm [Broken] works to block electromagnetic waves.
 
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SpaceTiger said:
The charge distribution on the inner surface would have to be non-uniform in order for there to be an electric field of zero inside the conducting shell. However, the distribution on the outer surface would be uniform. Conductors essentially "hide" the information about the charge inside them and that's why the http://www.absoluteastronomy.com/encyclopedia/f/fa/faraday_cage.htm [Broken] works to block electromagnetic waves.

college application approved :approve:

marlon

ps : it has been a while, man, how have you been ?
 
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SpaceTiger
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marlon said:
ps : it has been a while, man, how have you been
Pretty good, just chilling in Seattle for a few weeks. :biggrin:
 
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SpaceTiger said:
Pretty good, just chilling in Seattle for a few weeks. :biggrin:
what exactly does chilling mean ? :wink:

is it a grunge thing ?

marlon
 
SpaceTiger
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marlon said:
what exactly does chilling mean ? :wink:

is it a grunge thing ?
Heh, I was a grunge freak in high school, but no, it just means I'm relaxing. But I guess that's not entirely true, since I'm here to work with an old advisor. :tongue2:
 
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SpaceTiger said:
Uh, if I'm visualizing this correctly, it's the same from the outer surface to the inner surface (since you're inside a conductor), but inside the inner surface, the potential goes as 1/r.
ops! I type to quick that I didn't notice I write the wrong thing. You are right.
 

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