Simple charging by induction problem

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A positively charged sphere induces polarization in a grounded metal cup, resulting in a negative charge on the cup's outer surface and no charge on the inner surface. When the grounding wire is removed, the negative charges remain isolated on the cup. The excess electrons, pushed away from each other due to repulsion, settle on the exterior of the conductor to minimize energy. The discussion clarifies that "exterior" refers specifically to the outside surface, not the inside. Thus, the final charge distribution is negative on the outside and neutral on the inside.
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Homework Statement


A small positively charged sphere is lowered by a nonconducting thread into a grounded metal cup without touching the inside surface of the cup. The grounding wire attached to the outside surface is disconnected and the charged sphere is then removed from the cup. What best describes the subsequent distribution of excess charge on the surface of the cup?

The Attempt at a Solution


It's a multiple choice question and the answer is "Negative charge resides on the outside surface, and no charge resides on the inside surface."

I don't understand why no charge is in the inside surface though.

1. Sphere polarizes cup and ground. Cup is charged negative and ground is "positive".
2. Ground removed. Negative charges are isolated on cup.
3. Charges move to exterior surface of the conductor.

Doesn't the "exterior surface of the conductor" include the inside surface of the cup as well as the outside surface?
 
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A general rule: all net charge resides on the exterior of a conducting surface. Since the cup was grounded when the charged sphere was introduced, electrons from the outer surface were pulled to the inside of the sphere. "Replacement" electrons were pulled onto the cup through the grounding wire. Wire is removed, and charged sphere removed. There are now "excess electrons" on the cup. With no outside influence, those electrons will be pushing away from each other. This will cause them to reside on the outside, the furthest away they can get from each other, and thereby the lowest state of energy.

and "exterior" means "outside." Interior surface would be the inside surface. There is no ambiguity with the statement.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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