Polarization of a solid sphere of nonconducting material

  • Thread starter goohu
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  • #1
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Homework Statement:
see picture
Relevant Equations:
1) surface charge density = P * an
2) volume charge density = -\nabla \cdot P
3) Gauss law for polarization
a) Just using the equations gives us:
surface charge density = ## \rho_{\rho s} = kR^2 ##
volume charge density = ## \rho_\rho = -4kR ##

b) I am not sure here but the Q on the shell is the same as within. If thats the case we can use gauss law to find Q which I guess is the total charge.

## -Q = \int \rho_{\rho s} \cdot ds ##

My textbook states (for conductors) any introduced charge will move to the surface and redistribute itself due to repulsion. In this case the total charge on the shell is the same as "within"?

c) The E-field can also be found form gauss law. Then I assume you plug in 2a as R? This would be the standard way of solving the problem but IDK if nonconducting material is a special case?
 

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Answers and Replies

  • #2
TSny
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a) Just using the equations gives us:
surface charge density = ## \rho_{\rho s} = kR^2 ##
What is the value of ##R## at the surface?

volume charge density = ## \rho_\rho = -4kR ##
Looks good.

b) I am not sure here but the Q on the shell is the same as within. If thats the case we can use gauss law to find Q which I guess is the total charge.

## -Q = \int \rho_{\rho s} \cdot ds ##

My textbook states (for conductors) any introduced charge will move to the surface and redistribute itself due to repulsion. In this case the total charge on the shell is the same as "within"?

You should be able to calculate separately the total surface charge and the total volume charge. Then you can check to see of the total surface charge is equal and opposite to the total volume charge.

c) The E-field can also be found form gauss law. Then I assume you plug in 2a as R? This would be the standard way of solving the problem but IDK if nonconducting material is a special case?
Gauss' law is always valid.
 

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