# Conducting Shell

heman
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)??

## Answers and Replies

Homework Helper
Gold Member
All the charge on a conducting shell is on the outer surface.

heman
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.!

brasilr9
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.

Homework Helper
Gold Member
brasil is right. I misread the original question, and didn't realize there was a charge at r=0.

heman
That's okay Brasil but howcome,the same potential for outer surface becomes equal to that of inner surface.!

brasilr9
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.

marlon
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

brasilr9
Of course it isn't uniform

marlon
brasilr9 said:
Of course it isn't uniform

WRONG

marlon

Staff Emeritus
Gold Member
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.

Staff Emeritus
Gold Member
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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marlon
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

marlon

ps : it has been a while, man, how have you been ?

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Staff Emeritus
Gold Member
marlon said:
ps : it has been a while, man, how have you been

Pretty good, just chilling in Seattle for a few weeks.

marlon
SpaceTiger said:
Pretty good, just chilling in Seattle for a few weeks.
what exactly does chilling mean ?

is it a grunge thing ?

marlon

Staff Emeritus
Gold Member
marlon said:
what exactly does chilling mean ?

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:

brasilr9
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.