What happens when two spheres are connected by a conducting wire?

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Homework Equations
The Attempt at a Solution
Answers and Replies

Thread starter choco_moo
Start date Feb 18, 2010

In the figure, r1 = 5 and r2 = 7 cm. Before the conducting spheres are connected by the wire a charge of 5×10-7 C is placed on the smaller sphere while the larger sphere is uncharged. Calculate the charge on the smaller sphere after the wire is connected. Assume that the separation of the spheres is very large compared to their radii.

Homework Equations

V = kQ/r

The Attempt at a Solution

I know that at the beginning, only the smaller sphere has a charge. Therefore, the total charge of the system after being connected must equal the same charge. Also, I know that both spheres will have the same potential. V1 = V2 kQ1/r1 = kQ2/r2

After this point, I'm just clueless on how to figure out those individual charges knowing only the total charge and radii. Any help is great!

Answers and Replies

collinsmark

I think you've kind of conceptually solved the problem, even if you don't realize it yet. :tongue2: (The only other detail in the problem statement was "Assume that the separation of the spheres is very large compared to their radii," but I think that's just a way saying that you don't have worry about either sphere disrupting the spherical charge distribution of the other sphere. In other words, you should be able to use the normal equations for spheres.) Things that you have already given or said:

r1 and r2 are given in the problem statement. So they are not unknowns.

"the total charge of the system after being connected must equal the same charge." Yes, that's right, so Q1 + Q2 = 5×10-7 C. "Also, I know that both spheres will have the same potential. [...] kQ1/r1 = kQ2/r2"

So, by your own words, you are left with two equations, and two unknowns, ...

I kind of get what you're saying, but I still can't get it for some reason. This seems really simple, but I just can't figure out how I'd get Q1 when its set equal to another equation with another unknown (Q2).

collinsmark

I kind of get what you're saying, but I still can't get it for some reason. This seems really simple, but I just can't figure out how I'd get Q1 when its set equal to another equation with another unknown (Q2).

Well, you already know that Q1 + Q2 = 5×10-7 C. That means Q2 = (5×10-7 C) - Q1 Substitute that into kQ1/r1 = kQ2/r2. (Wherever it says "Q2" substitute "[5×10-7 C] - Q1")

Now you have 1 equation and 1 unknown. Solve for Q1. (r1 and r2 are known constants given in the problem statement, and you can find k in your textbook [its a fundamental constant])

I got it now. I was being so stupid. For some reason, I kept thinking I had to plug in (5e-7 - Q1) for Q2 and (5e-7 - Q2) for Q1 and had to solve for Q1 that way. Thanks for the help!

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There are two conducting spheres of different charge and a conducting wire. After they are connected by the wire, charge flows between the spheres. The charge distributes itself so that the spheres are at the same potential, but I have not been able to find an explanation for this. Why wouldn't the charge be distributed so that the electric field is the same, since that is what is moving the charges in the wire?