Charging the Small Sphere: Questions & Answers

[cragar]

I was reading this question out of a book. A small metal sphere hangs by an insulating thread within a larger hollow conducting sphere. A conducting wire extends from the small sphere through, but not touching, a small hole in the hollow sphere. A charged rod is used to transfer positive charge to the protruding wire.
And it wants to know what objects will be charged.
This is not a homework question just wondering.
SO just the inner smaller sphere will have positive charge on it, since the charge can't flow across the insulating thread. But the electric field inside a conductor is zero so the positive charge on the smaller sphere will attract negative charge on the inner shell of the larger sphere and then leave positive charges on the very outer shell of the sphere.
But what if the smaller sphere was hanging to the shell by a wire. Would the charge then just spread out uniformly to make the E field inside the conductors?
And then also why would the positive charge leave the rod to the metal sphere. Is the E field pushing the charge onto the sphere. And is the E field of the rod causing the metal sphere to be polarized and this attracts the positive charge from the rod?

 

[Antiphon]

If hanging by a wire, all the charge would go all the way around to the outside of the large sphere. There would be zero electric field in the hollow space.

When hanging by the insulator, there is a non-zero electric field between the spheres.

The positive charge leaves because the voltage on the rod is higher. This isn't given in the book but it is the case. The voltage is a function of the shape of the rod and the number of charges already on it. Q=CV as in a capacitor.

 

[Born2bwire]

Conductors are equipotential surfaces and the potential is dependent upon the surface charge that exists on the conductor. So if you have a charged conducting rod, then it will have a different voltage than your neutrally charged spheres and shells. However, when you bring the charged rod in contact with your small sphere then the charge must redistribute itself across the joined surfaces so that a new equipotential is reached.

As for a more physical picture, then yes I think you can take the induction of the charge separation on the small sphere to be the culprit. As you bring the positively charged rod into proximity of the sphere, it induce charge separation. The electrons will bunch up on the surface closest to the rod leaving positive ions on the opposite side. Upon touching, the electrons be pulled over by the positive ions on the rod thus depriving the sphere of some of its electrons resulting in a positive charge.

You can even see this easily enough yourself. Make a pith ball using a foil candy wrapper (gum or Ferrero Rocher) and wrap the foil around a piece of string. Then take a plastic comb and rub it against wool or cotton (sock does nicely). When you bring the comb into close proximity of the pith ball the charge induction can become so severe that the ball is physically attracted to the comb.

This does not necessarily mean that the charge is evenly distributed. We can make that claim in some cases by virtue of symmetry though. For example, if you connect the inner and outer sphere by a conducting wire as you propose then the charge distribution should be symmetric about the axis of the wire (say along the \phi direction) but it can vary along the \theta direction due to the breaking of symmetry.

 

[cragar]

Ill have to get some gum and a string and make that. thanks for both of your answers.

 

[pmacias]

I can provide some insights into the questions you have raised. First, let's discuss the charging of the small sphere. When a charged rod is brought near the protruding wire, the charge on the rod will induce a separation of charges on the wire. The end closest to the rod will acquire a negative charge, while the end attached to the small sphere will acquire a positive charge. This positive charge will then transfer to the small sphere, as the electric field of the charged rod exerts a force on the positive charges, causing them to move onto the sphere.

Now, let's consider the scenario where the smaller sphere is hanging by a wire. In this case, the charge will spread out uniformly on the surface of the smaller sphere and the wire, as the electric field inside a conductor is zero. This is known as electrostatic shielding, where the conductor shields the inside from external electric fields.

As for why the positive charge will leave the rod and transfer to the metal sphere, it is due to the electric field created by the charged rod. This electric field exerts a force on the positive charges, causing them to move towards the sphere. And yes, the electric field of the rod does cause the metal sphere to be polarized, as the positive charges on the sphere will align themselves in response to the electric field of the rod.

I hope this helps to clarify your questions. It's important to understand the role of electric fields in the transfer and distribution of charges in conductors. Keep exploring and asking questions - that's what scientists do!