Electric field near conducting shell

  • #1
lys04
51
3
Homework Statement
Electric field near conducting shell
Relevant Equations
E=kQ/r^2
How would I do this question? I am having trouble figuring out what the radius is meant to be, why is it not 3R?
1691475588892.png
 
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  • #2
The charge is inside a conducting shell. What is the relevance of that?
 
  • #3
PeroK said:
The charge is inside a conducting shell. What is the relevance of that?
Induces a charge in the inner surface of the shell?
 
  • #4
lys04 said:
Induces a charge in the inner surface of the shell?
Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?
 
  • #5
PeroK said:
Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?
I've learnt Gauss's law, if that's applicable.
 
  • #6
lys04 said:
I've learnt Gauss's law, if that's applicable.
Gauss's law is useful. What can you say about the potential on the surface of a conductor?
 
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  • #7
lys04 said:
I've learnt Gauss's law, if that's applicable.
In a way: you will have to argue why the required symmetry (*) is present.

spherical symmetry of the field outside the shell

##\ ##
 
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  • #8
PeroK said:
Gauss's law is useful. What can you say about the potential on the surface of a conductor?
It's uniform?
 
  • #9
BvU said:
In a way: you will have to argue why the required symmetry (*) is present.

spherical symmetry of the field outside the shell

##\ ##
Yeah I'm not sure about that
 
  • #10
lys04 said:
It's uniform?
Yes. The surface of the shell must be an equipotential. Now, the potential is fully determined by the boundary conditions (technically this is a uniqueness property of electrostatic solutions). You know that outside the shell a spherically symmetric potential is a solution, so it must be the only solution.

Now, you can apply Gauss's law.
 
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