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Charging a spherical shell by conduction

by bigplanet401
Tags: charging, conduction, shell, spherical
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bigplanet401
#1
Aug27-12, 07:39 PM
P: 67
1. The problem statement, all variables and given/known data

An insulated spherical conductor of radius R1 carries a charge Q. A second conducting sphere of radius R2 and initially uncharged is then connected to the first by a long conducting wire.

(a) After the connection, what can you say about the electric potential of each sphere?
(b) How much charge is transferred to the second sphere?
(c) The spheres are assumed to be far apart compared to their radii. Why make this assumption?

2. Relevant equations



3. The attempt at a solution

(a) Once the spheres are "hooked up," they behave as one conductor. Because a conductor is an equipotential, the potential of each sphere is the same.

(b) Because the potentials have to be the same, the charge on each sphere is in
proportion to their radii:
[tex]
V = \frac{Q-\Delta Q}{R_1} = \frac{\Delta Q}{R_2}
[/tex]

Delta Q is the amount of charge exchanged between the spheres, and the total charge
should be Q because of conservation of charge. The final charges are then
[tex]
Q_1 = Q \frac{R_1}{R_1+R_2} \quad Q_2 = Q \frac{R_2}{R_1+R_2}
[/tex]

(c) No idea here. As the second sphere gains charge, shouldn't there be repulsion between the two (proportional to the inverse square of the separation distance)? But then what happens? I can't understand why the distance would matter.
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TSny
#2
Aug27-12, 08:10 PM
HW Helper
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P: 4,821
There are certain conditions which must be met for the equation V = Q/R to be valid for the potential at the surface of a charged sphere. For example, think about how the charge on the sphere must be distributed in order for the equation to be valid.

[Your work on parts (a) and (b) looks very good.]
bigplanet401
#3
Aug27-12, 08:38 PM
P: 67
The charge has to be uniform, and after a certain amount of time, the spheres would each have net charge. If they were brought close to one another, the net charge would be "pushed away" from the point where they're closest. So then the potential wouldn't be spherically symmetric. Is this right?

What about the two sides that are closest, though. Is there negative charge (the wire is still connected), or is this area neutral? My guess is

(+Qnet neutral)<-- d -->(neutral +Qnet)

...but could it be (with Q-q = Qnet)

(+Q -q)<-- d -->(-q +Q)

instead?

TSny
#4
Aug27-12, 09:16 PM
HW Helper
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P: 4,821
Charging a spherical shell by conduction

Quote Quote by bigplanet401 View Post
The charge has to be uniform, and after a certain amount of time, the spheres would each have net charge. If they were brought close to one another, the net charge would be "pushed away" from the point where they're closest. So then the potential wouldn't be spherically symmetric. Is this right?
Yes, the charge density on the surface of the sphere has to be uniform and the sphere must be isolated for V = Q/R to be valid.
What about the two sides that are closest, though. Is there negative charge (the wire is still connected), or is this area neutral? My guess is

(+Qnet neutral)<-- d -->(neutral +Qnet)

...but could it be (with Q-q = Qnet)

(+Q -q)<-- d -->(-q +Q)

instead?
As long as the spheres remain connected by a wire, I believe the charge density on the surfaces of the spheres would remain positive for all regions of the surfaces, even on the sides that are closest. But, the charge density would be more concentrated on the far sides.

The two spheres connected by a wire make one overall isolated conductor. I think you can construct an argument for showing that no matter what the shape of an isolated conductor, if you put positive charge on the conductor then there will be no regions of the surface that would have a negative charge density. But I would enjoy seeing a counterexample or counterargument.

For conducting spheres that are not connected to each other, an interesting thing can happen. See http://www.nature.com/news/like-attracts-like-1.10698
bigplanet401
#5
Aug27-12, 09:47 PM
P: 67
Thanks for your help!


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