Charge on spherical conducting shells

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Homework Help Overview

The discussion revolves around the electrostatics of two spherical conducting shells, one inside the other, with specified charges on each shell. The original poster seeks to determine the charge distribution on the inner and outer surfaces of both shells given the total charges of -3q and +5q.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of charge distribution on conducting shells, questioning the assumptions about charge movement and the application of Gauss's law. There is discussion about how charges would redistribute and the reasoning behind the proposed charge values on the surfaces.

Discussion Status

Participants are actively engaging with the problem, raising questions about the original poster's assumptions and reasoning. Some guidance has been offered regarding the application of Gauss's law and the behavior of charges within conductors, but no consensus has been reached on the final charge distribution.

Contextual Notes

There is a mention of the shells being charged and not neutral or grounded, which may influence the charge distribution. The discussion includes confusion regarding the relationship between charges on different surfaces of the conductors.

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Homework Statement

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There are two spherical conducting shells one inside the other. If the total charge on the inside shell is -3q and the total charge on the outside shell is +5q what's is the charge on the inner and outer surface of each shell?

Attempt at solution

My teacher said i can assume there is no charge in the center so i think the inner surface of the small shell has a charge of -6q and the outer surface +3q and the larger shell has inner surface charge of -3q and outer surface charge of +8q
 
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Hello TP,

I grant you that this isn't about relevant equations. So the 2 from the template can be left out. But thinking up a solution isn't physics. What are your considerations to come to e.g. this -6 q ?
 
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Since there is no charge in the center i wanted to use the smallest numbers that could make the net charge cancel to negative three. I know inside the conductor must be 0 so the charge is on the surfaces so i picked -6 for the inside surface and positive 3 for the outside
 
Wouldn't the -6 q race to the outer surface in a flash ? All the little charge carriers repel each other like crazy and they want to sit as far away from each other as possible !

Remember that inside a conductor there is no electric field: if there were, the charge carriers would simply move with the field until either the field is zero or they can't go any further !

Do you know about Gauss's law ?
 
the internmost charge is the same as the one on the outer surface?
 
Still don't know if you know about Gauss's law. Let's assume you do. Else look it up, e.g. on hyperphysics.

Imagine a gauss sphere halfway inner and outer surfaces of the inner shell. No field (it's a conductor) hence no (net) charge within. No charge inside the inner surface, so no charge on the inner surface either. Conclusion: on the outer surface of the inner shell there is sitting -3q .

Now it's your turn to find out how much charge is sitting on the inner surface of the outer shell ...
 
Oops sorry I do know gausses law and I know the electric field would be E = Qencl/ε4πr^2 it is the Qencl I am having trouble with.
I was confused because I though that if there was charge on one surface then there must be charge on the other one. If the only charge on the inner sphere is -3q on the outer surface then the inner surface of the outer sphere would have a +3q charge and the outer surface of the outer sphere would have a + 2q charge. this also confused me when i was trying the problem on my own because I thought the same polarity of charge could not be on both surfaces of the conductor
 
The shells are prepared by charging (they are not neutral and not connected to ground). The distribution is as you now describe: the +3 is pulled inwards. That way a Gauss suface halfway inner and outer surfaces of the outer shell encloses zero Coulomb: no flux, no E field inside that conductor. The remaining +2 is distributed evenly over the outer surface of the outer shell (*). The outside world observes +2, e.g. using Gauss. (and has no way of telling what's inside !).(*) even if the inner shell isn't centered wrt the outer. Clear from Gauss's law and the outer shell being a conductor !
 
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thank you!
 

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