Potential difference of concentric conducting shells

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SUMMARY

The discussion focuses on calculating the potential difference between two concentric conducting spherical shells with charges +q and -q. To find the potential difference V_{a}-V_{b}, participants emphasize using Gauss's Law to determine the electric field E between the shells and integrating it to find the potential difference. The charge Q in Gauss's Law represents the charge enclosed by the Gaussian surface, which must be correctly identified to compute E. The integration bounds for calculating the potential difference can be chosen arbitrarily, as the sign of the result will adjust accordingly.

PREREQUISITES
  • Understanding of Gauss's Law and its application in electrostatics
  • Familiarity with electric fields and potential difference concepts
  • Basic calculus skills for performing integration
  • Knowledge of spherical coordinates and their application in physics
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Learn how to calculate electric fields for spherical charge distributions
  • Explore the concept of electric potential and its relation to electric fields
  • Practice integration techniques for calculating potential differences in electrostatics
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Students in physics, particularly those studying electromagnetism, as well as educators and anyone looking to deepen their understanding of electric fields and potential differences in concentric conducting shells.

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

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

\Delta V = -\int\stackrel{\rightarrow}{E}dl

My question is, using Gauss's Law, (\oint E_{n}dA = \frac{Q}{\epsilon}), 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
    37.1 KB · Views: 933
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: 1,187
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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