Potential difference of concentric conducting shells

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Homework Help Overview

The problem involves calculating the potential difference between two concentric conducting spherical shells with equal and opposite charges. The inner shell has a positive charge, while the outer shell has a negative charge. The task requires using Gauss's Law and integrating the electric field to find the potential difference.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Gauss's Law to determine the charge enclosed by a Gaussian surface and how to find the electric field in the region between the shells. Questions arise regarding the appropriate placement of the Gaussian surface and the integration bounds for calculating the potential difference.

Discussion Status

Participants are actively exploring the problem, with some providing hints and guidance on how to approach the integration of the electric field. There is an ongoing discussion about the significance of the Gaussian surface and the implications of integration limits on the final result.

Contextual Notes

There is a focus on understanding the relationship between the electric field and potential difference, as well as the implications of charge distribution in the context of the problem. Participants are encouraged to clarify their assumptions and reasoning throughout the discussion.

faint545
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Two concentric conducting spherical shells have equal and opposite charges. The inner shell has outer radius [itex]a[/itex] and charge [itex]+q[/itex]; the outer shell has inner radius [itex]b[/itex] and charge [itex]-q[/itex]. Find the potential difference [itex]V_{a}-V_{b}[/itex] between the shells

My professor said to in order to solve this, integrate the electric field to find the electric potential...

[itex]\Delta V = -\int\stackrel{\rightarrow}{E}dl[/itex]

My question is, using Gauss's Law, ([itex]\oint E_{n}dA = \frac{Q}{\epsilon}[/itex]), how do I go about finding Q?

Isn't Q just the charge of the shell?
 
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In Gauss's law, Q is the charge enclosed by your Gaussian surface. So, first decide where you Gaussian surface will be, then add up all the charge inside of it.

HINT: You want your surface to be in the region where you want to find the electric field.
 
G01 said:
In Gauss's law, Q is the charge enclosed by your Gaussian surface. So, first decide where you Gaussian surface will be, then add up all the charge inside of it.

HINT: You want your surface to be in the region where you want to find the electric field.

This is what I have drawn (see attachment). Is the basic idea here to integrate the electric field of the outer Gaussian surface from b to a? If so, what about the inner Gaussian surface?
 

Attachments

  • 2011-09-27_20-21-30_086.jpg
    2011-09-27_20-21-30_086.jpg
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You want to find the potential difference between the shells, so you don't need the Gaussian surface outside the larger shell.

Try this: Take your inner surface and place it at an arbitrary point r. Then, find E using the standard approach when using Gauss's law. You will then have E between the plates as a function of r. Can you use that to find the potential difference between the plates?
 
G01 said:
You want to find the potential difference between the shells, so you don't need the Gaussian surface outside the larger shell.

Try this: Take your inner surface and place it at an arbitrary point r. Then, find E using the standard approach when using Gauss's law. You will then have E between the plates as a function of r. Can you use that to find the potential difference between the plates?

Alright... see attached image.

Now, how do i determine if i should integrate from a to b or from b to a?
 

Attachments

  • 2011-09-28_12-25-07_614.jpg
    2011-09-28_12-25-07_614.jpg
    47.3 KB · Views: 1,196
faint545 said:
Alright... see attached image.

Now, how do i determine if i should integrate from a to b or from b to a?

That all looks good.

As far as your integration bounds are concerned, it doesn't matter. The sign of your end result will change, but that's just like hooking up a voltmeter in reverse: You will still get the right pot. difference, just the sign will change.
 
thanks for your guidance
 
faint545 said:
thanks for your guidance

No problem! :smile:
 

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