Two Conducting Spheres connected by a wire

AI Thread Summary
Two conducting spheres of different radii connected by a wire will reach an equilibrium charge distribution, but the surface charge densities will not be equal due to their differing sizes. The charge on sphere B can be calculated as (rB/rA) * Qa, indicating that the charge distribution is dependent on the radii of the spheres. The electric field strength at the surface of sphere A is greater than that of sphere B, which contradicts the assumption of equal surface charge densities. This difference arises because excess charge accumulates more on regions with smaller radii of curvature, leading to varying electric field strengths. Understanding these principles clarifies the behavior of charges on conductors in electrostatic equilibrium.
tomizzo
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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 = \eta/\epsilonnaught

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\pi*rA^2)=Qb/(4\pi*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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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 = \eta/\epsilonnaught

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\pi*rA^2)=Qb/(4\pi*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?
 
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?
 
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.
 
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?
 
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.
 
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
 
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