- #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?
Induces a charge in the inner surface of the shell?PeroK said:The charge is inside a conducting shell. What is the relevance of that?
Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?lys04 said:Induces a charge in the inner surface of the shell?
I've learnt Gauss's law, if that's applicable.PeroK said:Have you studied that? Or, learned any techniques to find the electric field outside an asymmetric charge configuration?
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.lys04 said:I've learnt Gauss's law, if that's applicable.
It's uniform?PeroK said:Gauss's law is useful. What can you say about the potential on the surface of a conductor?
Yeah I'm not sure about thatBvU said: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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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.lys04 said:It's uniform?