Electric field near conducting shell

AI Thread Summary
The discussion centers on understanding the electric field near a conducting shell with a charge inside. Participants express confusion about the radius and the implications of the charge being within the shell, particularly regarding induced charges on the inner surface. Gauss's law is highlighted as a key tool for analyzing the electric field, emphasizing the importance of symmetry in the problem. The surface of the conducting shell is confirmed to be an equipotential, and the potential outside the shell is determined by boundary conditions. Overall, applying Gauss's law is essential for solving the problem effectively.
lys04
Messages
144
Reaction score
5
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
 
Physics news on Phys.org
The charge is inside a conducting shell. What is the relevance of that?
 
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?
 
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?
 
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.
 
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?
 
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

##\ ##
 
PeroK said:
Gauss's law is useful. What can you say about the potential on the surface of a conductor?
It's uniform?
 
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.
 
Back
Top