Electric field inside 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
1 reply · 2K views
bigplanet401
Messages
101
Reaction score
0

Homework Statement



A spherical conducting shell of inner radius a and outer radius b contains a centrally-located point charge +Q. Is an electric field present at radii (i) less than a, (ii) between a and b, and (iii) greater than b?

Homework Equations





The Attempt at a Solution



My textbook says the electric field inside a conductor (a<r<b here) is zero, but I can't understand why. In (i), the field is that due to the charge. In (iii), field lines emanate from the surface at right angles (because the induced charge is positive on the outer surface).

In (ii), it seems like the field should be non-zero because of the induced positive charge on the outer surface and the induced negative charge on the inner surface. Wouldn't this create an electric field directed radially inward from r=b to r=a? The textbook says it's zero.

Thanks for your help!
 
Physics news on Phys.org
bigplanet401 said:

Homework Statement



A spherical conducting shell of inner radius a and outer radius b contains a centrally-located point charge +Q. Is an electric field present at radii (i) less than a, (ii) between a and b, and (iii) greater than b?

Homework Equations





The Attempt at a Solution



My textbook says the electric field inside a conductor (a<r<b here) is zero, but I can't understand why. In (i), the field is that due to the charge. In (iii), field lines emanate from the surface at right angles (because the induced charge is positive on the outer surface).

In (ii), it seems like the field should be non-zero because of the induced positive charge on the outer surface and the induced negative charge on the inner surface. Wouldn't this create an electric field directed radially inward from r=b to r=a? The textbook says it's zero.

Thanks for your help!

Yes, the induced negative charge on the inner surface and the induced positive charge on the outer surface create a radially inward electric field that is just enough to cancel the radially outward electric field from the central point charge in this region a < r < b.

The point of an ideal conductor is that it can be considered a sort of limitless source of free charge. So, the E-field from the central point charge pushes charges around inside the conductor until they are arranged in such as way as to cancel out the field. If the field were not canceled out, and there were still a NET electric field inside the conductor, then this net electric field would push around even MORE charges until the arrangement was such that the field inside the conductor was zero.