Two Conducting Spheres Connected by a Wire

In summary, two conducting spheres with radii of 5 and 7 cm are connected by a wire. Before the connection, the smaller sphere has a charge of 5×10-7 C while the larger sphere is uncharged. After the connection, the total charge of the system remains the same, and both spheres have the same potential. Using the equation for potential, it can be determined that the charge on the smaller sphere after the connection is 5×10-7 C - Q1, where Q1 is the charge on the larger sphere. Solving for Q1 gives the final answer.
  • #1
choco_moo
26
0

Homework Statement


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.

2wcn29s.jpg



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!
 
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  • #2
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, ...
 
  • #3
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).
 
  • #4
choco_moo said:
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])
 
  • #5
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!:smile:
 

1. How are the charges distributed on the two conducting spheres?

The charges on the two conducting spheres are distributed in such a way that they are equal in magnitude and opposite in sign. This is due to the flow of electrons from one sphere to the other through the connecting wire, causing both spheres to have an equal and opposite charge.

2. How does the distance between the two spheres affect the charge distribution?

The charge distribution on the two spheres is affected by the distance between them. As the distance decreases, the electric field strength increases, causing a greater flow of electrons between the spheres and resulting in a higher charge on each sphere.

3. What happens to the charge on the two spheres if the wire connecting them is removed?

If the wire connecting the two spheres is removed, the charges on the spheres will remain the same. This is because the charges have already redistributed to reach equilibrium between the two spheres. However, if the spheres are then separated, the charges will remain on each sphere, but the total charge on each sphere will decrease due to the loss of electrons through the wire.

4. Can the charge distribution on the spheres be changed by an external electric field?

Yes, the charge distribution on the spheres can be changed by an external electric field. If the external field is stronger than the electric field between the two spheres, it can cause the charges to redistribute and result in an unequal charge distribution on the spheres. This effect is known as induction.

5. How does the material of the spheres affect the charge distribution?

The material of the spheres does not affect the charge distribution, as long as they are both conductors. Conductors allow charges to flow freely, so the charges will distribute in the same manner regardless of the material. However, if one of the spheres is a non-conductor, such as an insulator, the charges will not be able to flow between the spheres and the charge distribution will be different.

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