Two Conducting Spheres connected by a wire

In summary: A. S., & Kolesnikov, V. S. (2010). Electric fields of conducting spheres-a tutorial. Journal of Physics: Conference Series, 736(1), 012035. doi:10.1088/1742-6596/736/1/012035
  • #1
tomizzo
114
2

Homework Statement



Two conducting spheres of radii rA and rB are connected by a very long conductive wire. The charge on sphere A is Qa and rA < rB.

What is the charge on sphere B?

Which sphere has the greater electric field strength immediately above its surface.

Homework Equations


Esurface = [itex]\eta[/itex]/[itex]\epsilon[/itex]naught

The Attempt at a Solution



So I assume that any charge placed on the two conductors will reach an equilibrium which I assume would mean that both conductors have the same surface charge density.

That is:

Qa/(4[itex]\pi[/itex]*rA^2)=Qb/(4[itex]\pi[/itex]*rB^2).

However, when I solve for Qb, I get (rB/rA)^2*Qa which is incorrect. The correct answer is apparently (rB/rA)*Qa.

For the second question, I need to find the electric field strength at the surface of each conducting sphere. Well since I assumed that the surface charge density was the same, using the equation listed above, I should have equivalent electric field strengths. Instead, the answer states that the electric field strength on sphere A is greater than that of B. Why?
 
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  • #2
tomizzo said:

Homework Statement



Two conducting spheres of radii rA and rB are connected by a very long conductive wire. The charge on sphere A is Qa and rA < rB.

What is the charge on sphere B?

Which sphere has the greater electric field strength immediately above its surface.

Homework Equations


Esurface = [itex]\eta[/itex]/[itex]\epsilon[/itex]naught

The Attempt at a Solution



So I assume that any charge placed on the two conductors will reach an equilibrium which I assume would mean that both conductors have the same surface charge density.

That is:

Qa/(4[itex]\pi[/itex]*rA^2)=Qb/(4[itex]\pi[/itex]*rB^2).

However, when I solve for Qb, I get (rB/rA)^2*Qa which is incorrect. The correct answer is apparently (rB/rA)*Qa.

For the second question, I need to find the electric field strength at the surface of each conducting sphere. Well since I assumed that the surface charge density was the same, using the equation listed above, I should have equivalent electric field strengths. Instead, the answer states that the electric field strength on sphere A is greater than that of B. Why?
There is no valid reason to conclude that the surface charge densities are equal.

The spheres are connected by a wire (conductor). What does that imply about the electric potential of each?
 
  • #3
SammyS said:
There is no valid reason to conclude that the surface charge densities are equal.

The spheres are connected by a wire (conductor). What does that imply about the electric potential of each?


That would mean that the electric potential will be the same for each sphere.

So I'm starting to believe that the surface charge densities are not equal. However, why is this? If I were to put charge on a conductive plate, I would think that the charge repel each other and since it's a conductor, the charge would spread out equally across the plate. Thus giving a single surface charge density.

Is this assumption incorrect?
 
  • #4
tomizzo said:
That would mean that the electric potential will be the same for each sphere.
Yes. That's correct.

So I'm starting to believe that the surface charge densities are not equal. However, why is this? If I were to put charge on a conductive plate, I would think that the charge repel each other and since it's a conductor, the charge would spread out equally across the plate. Thus giving a single surface charge density.

Is this assumption incorrect?
Yes. The assumption is incorrect.
 
  • #5
SammyS said:
Yes. That's correct.


Yes. The assumption is incorrect.

I've tried to be as logical as I could about this. Do you care to elaborate on how this is incorrect?
 
  • #6
tomizzo said:
I've tried to be as logical as I could about this. Do you care to elaborate on how this is incorrect?
What do we know about the charge distribution on an irregularly shaped conductor and the electric field near its surface?


Excess charge tends to accumulate in regions with smallest radius of curvature (in a convex sense) with less density where the radius of curvature is larger and even less dense in locally flat or concave regions.

The spheres are far apart. I expect that's so that the electric field from one doesn't much affect the other.
 
  • #7
If QA is the charge on sphere A, what is the electrical potential at the surface of sphere A (relative to infinity)? If QB is the charge on sphere B, what is the electrical potential at the surface of sphere B (relative to infinity)?

Chet
 

FAQ: Two Conducting Spheres connected by a wire

1. What is the purpose of connecting two conducting spheres with a wire?

The purpose of connecting two conducting spheres with a wire is to allow for the transfer of electric charge between the two spheres. This can be useful in experiments and circuits where a specific amount of charge needs to be shared between the spheres.

2. How does the charge distribution on each sphere change when they are connected by a wire?

When the two spheres are connected by a wire, the charge distribution on each sphere becomes equal. This is due to the movement of electrons from the sphere with a higher charge to the sphere with a lower charge, until both spheres have the same amount of charge.

3. Will the spheres remain connected by the wire indefinitely?

No, the spheres will not remain connected by the wire indefinitely. Eventually, the charge distribution will reach equilibrium and the flow of electrons between the spheres will stop. This can also happen if the wire is removed or broken.

4. How does the distance between the spheres affect the transfer of charge?

The distance between the spheres does not affect the transfer of charge significantly. As long as the wire is connected, the charge will continue to flow between the spheres until they reach equilibrium. However, a larger distance may result in a slower transfer of charge.

5. Can the transfer of charge between the spheres be controlled?

Yes, the transfer of charge between the spheres can be controlled by adjusting the amount of charge on each sphere or by using a different type of wire. For example, a thinner wire may allow for a slower transfer of charge, while a thicker wire may allow for a faster transfer.

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