Charged Metallic Sphere Touching Spherical Shell From Inside

[chayanne]

Homework Statement

(From Physics for Scientists and Engineers, 7E, Serway-Jewett Chapter 25 Q11)

(i) A metallic sphere A of radius 1 cm is several centimeters away from a metallic spherical shell B of radius 2 cm. Charge 450 nC is placed on A, with no charge on B or anywhere nearby. Next, the two objects are joined by a long, thin, metallic wire (as shown in Fig. 25.20), and finally the wire is removed. How is the charge shared between A and B?

(a) 0 on A, 450 nC on B
(b) 50 nC on A and 400 nC on B, with equal volume charge densities
(c) 90 nC on A and 360 nC on B, with equal surface charge densities
(d) 150 nC on A and 300 nC on B
(e) 225 nC on A and 225 nC on B
(f) 450 nC on A and 0 on B
(g) in some other predictable way
(h) in some unpredictable way

(ii) A metallic sphere A of radius 1 cm with charge 450 nC hangs on an insulating thread inside an uncharged thin metallic spherical shell B of radius 2 cm. Next, A is made temporarily to touch the inner surface of B. How is the charge then shared between them? Choose from the same possibilities. Arnold Arons, the only physics teacher yet to have his picture on the cover of Time magazine, suggested the idea for this question.

Homework Equations

- V=kq/r (used in part (i))

Possibly relevant :
- V=k ∫dq/r,
- V=-∫E°dl,

The Attempt at a Solution

For part (i) I was able to reach an answer using V₁ = V₂, assuming conservation of charge:

kq₁/r₁ = kq₂/r₂;
where q₁+q₂=450 nC

However for part (ii) I am pretty lost. Ideas I have tried were a similar solution to part (i), a cavity within a conductor (suggesting E inside the spherical shell should be 0), or the fact that net electric charge of a conductor is on its surface.

My attempts were unfruitful. A similar solution to (i) doesn't make all that sense to me, as I feel like they become the same thing when they touch one another (sounds romantic). Cavity within a conductor (which I literally just pulled off the textbook as I was looking for an idea to help me solve it) kind of seemed to be helpful, however, I might be misinterpreting it to make my case here. The net electric charge on surface idea also seems to lose ground when I think about the inside sphere's surface being part of the "sphere inside shell" object's surface, therefore having charge on it.

Thank you very much for your help and time.

 

[mfb]

or the fact that net electric charge of a conductor is on its surface.

That is sufficient. When they touch, you can consider them as a single conductor. Where is the whole charge? It is important that the outer sphere has no hole here.

 

[chayanne]

Thank you very much for the response. The whole charge is on the surface, but doesn't that include the inner surface and thus the surface of the inner sphere? Why does it lose the surface charge when it goes back to its original position? Thanks again.

 

[mfb]

An inner surface with charge would lead to a field inside. That cannot happen - inner surfaces are always regions of constant potential, and there is nowhere the electric field lines could end (no additional charges separated from your object).

 

[Nickie]

The solution to part (i) is (f) 450 nC on A and 0 on B. This is because when the two objects are joined by a wire, they become a single conducting object with a total charge of 450 nC. Since they are both conductors, the charge will distribute itself evenly on the surface of the combined object, resulting in 450 nC on A and 0 on B.

For part (ii), after the two objects are touched, they become a single conductor with a total charge of 450 nC. However, since the inner surface of the spherical shell is now in contact with the inner sphere, the charge will distribute itself evenly on the combined surface of the inner sphere and the inner surface of the shell. This means that the charge will be shared between the two surfaces, resulting in a different distribution than in part (i). The solution is (c) 90 nC on A and 360 nC on B, with equal surface charge densities. This can be calculated using the equation for electric potential, V=k ∫dq/r, and integrating over the surface of the combined object.

I hope this helps clarify the solution for you. It's important to remember that when conductors are in contact, they will redistribute their charge to achieve a state of equilibrium, with the charge evenly distributed on their surfaces.