Gauss' Law and charges placed within a spherical conductor

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
accidentprone
Messages
16
Reaction score
0

Homework Statement



A hollow sherical conducting shell is suspended in air by an insulated string. The total charge on the conductor is -6 microCoulombs. If an additional point charge of +2 microCoulombs is placed in the hollow region inside the shell what is the total charge induced on the inside surface of the shell.


Homework Equations





The Attempt at a Solution


I know already that a charge will be induce on the inside surface as a closed, hollow conductor shields its interior from fields due to charges outside, but does not shield
its exterior from fields due to charges placed inside it. However is the answetr simply that -2 microCoulombs will be induced on the inside of the sphere? Or will I have to account for the charge on the outside also interacting with the charge induced on the inside surface?
Many thanks in advance!
 
on Phys.org
accidentprone said:
However is the answetr simply that -2 microCoulombs will be induced on the inside of the sphere?
Yes, that's all there is to it.
Or will I have to account for the charge on the outside also interacting with the charge induced on the inside surface?
No. They do not affect the inside surface charges.
 
Thank you!

One last question. I'm now asked to find the electric field immediately outside the conductor, given that the outside surface has a radius of 8 cm. Will The electric field outside be only produced by the charges on the outside surface as the inside and outside charges do not interact?
 
accidentprone said:
Will The electric field outside be only produced by the charges on the outside surface as the inside and outside charges do not interact?
Effectively, yes. But electric field is always produced by all the charges--it's just that the effect of the charges inside the outer surface is zero: They cancel out.
 
Thank you again, especially for the speedy replies - much appreciated!