# Homework Help: Unusual electric field

1. Mar 8, 2004

### frznfire219

This question is a bit strange, and I would appreciate help:

A solid uncharged conducting sphere has radius 3a and
contains a hollowed spherical region of radius 2a. A point
charge +Q is placed at a position a distance a from the
common center of the spheres. What is the magnitude of the
electric field at the position r = 4a from the center of the
spheres as marked in the figure by P?

(diagram at http://www.compadre.org/psrc/evals/Physics_Bowl_2003.pdf page 13 if it helps)

So apparently the field lines must emerge perpendicularly from the conducting sphere, so they would converge at the center. But that's not very satisfying for an explanation.

Any help is appreciated.

2. Mar 9, 2004

### paul11273

I am going to say that the spheres are only there to misdirect you. Since they have no charge, and they are non conducting, they should have no effect on the electric field generated by the point charge.

That said, just calculate the E-field at point P like you normally would for a point charge.

E=k(Q/r^2)

Fill in your r for this problem and you have the answer.

Anyone else have a take on this?

3. Mar 9, 2004

### Staff: Mentor

conductor as a shield

The sphere is a conductor. That's essential. Now consider the fact that the field inside a conductor must be zero.

4. Mar 9, 2004

### paul11273

Sorry, for some reason I read it as the sphere being non-conductive, but the question clearly states otherwise. I don't know what I was thinking.

So, can we use Gauss's Law on this, with a sphere centered at point Q with a radius of 3a?

Would the answer for this work out to E=k(Q/r^2) since q(induced)=Q, or would that only work if the point charge were at the center of the conducting hollow sphere?

I think that although the point charge is off center, the induced charge on the surface of the conducting sphere would be distributed uniformly.

5. Mar 9, 2004

### Staff: Mentor

A Gaussian surface that cuts through a conductor? Yikes.
Don't know how you've defined "r". But, using a Gaussian surface through the conductor tells you that the charge on the inner surface must equal -Q.
Now you've got it. The induced charge on the outer surface will distribute uniformly.

6. Mar 9, 2004

### frznfire219

i thought that the field must be uniform to be able to easily use gauss's law.

i'm not seeing why is this true?

7. Mar 10, 2004

### sickboy

It is so, because in electrostatic equilibrium there electric field inside the conducting volume is exactly zero. The external field causes the free electrons to accumulate in the surface of the conductor. This causes an internal field of an exactly opposite direction to the external field. In equilibrium, the fields are of the same magnitude. Thus, the net electric field inside a conducting volume is exactly zero.

Why can the charge only be on the surface, then? Consider a small volume inside the volume of the conductor. Since E=0 everywhere inside the conductor, the left side of Gauss' law becomes zero, which of course means that the right side, $\frac{Q}{\epsilon_o}$, also must vanish. This means that, no matter how the volume segment dV is chosen, there can't any charge inside it. Thus, there can't be any charge inside the conducting volume - only place where the charge can be is the on the surface of the conductor.

8. Mar 11, 2004

### Staff: Mentor

I'm trying to come up with a general argument showing this (that the charge on the sphere will distribute uniformly despite an uncentered charge within the cavity) but I'm having trouble. Perhaps that part of my brain has finally died! I'll keep working on it; I have not forgotten.

When there is no charge in the cavity, or when the charge is at the center, you will agree that the charge on the surface will distribute uniformly. Right?

9. Mar 11, 2004

### frznfire219

yes, definitely.

10. Mar 11, 2004

Why wouldn't it be true? Once the inner surface of the conductor arranges its charges to cancel the charge inside the inner surface, there's no electric field due to the inner surface and charge inside, but there is extra charge left over in the volume of the conductor. Naturally, all of this charge will distribute itself uniformly over the outer surface of the conductor and will not at all be affected by the inner surface and inner charge because the two cancel each other!

11. Mar 12, 2004

### Staff: Mentor

I certainly agree that this is true. But I quickly realized that I couldn't prove that the charge on the inner surface of the cavity exactly cancels the field due to the charge inside the cavity. All I could say is that the net effect of all the charges---inside the cavity, on the wall of the cavity, and on the the outer surface of the conductor---adds up to zero field inside the conducting material. A much weaker statement.

I have since proven that indeed the field due to the charges on and in the cavity does exactly cancel for all points outside the cavity. And that the charge distribution on the outside of the conductor is independent of the arrangement of charges within the cavity. But the proof, while simple, uses boundary conditions and uniqueness theorems. Do you have a simple argument that shows this? (There must be one that I am missing.)

In any case, for frznfire219, there's another simple way of looking at this particular problem since the conductor is spherical: Since the electric field must be perpendicular to the surface of a conductor, and all lines perpendicular to the surface of a sphere pass through the center, then it is reasonable that the field outside a spherical conductor must be equivalent to that of a charge placed at its center.

12. Mar 12, 2004

The argument is pretty much that charge inside the conductor is going to keep moving because of the cavity's electric field until that field goes to zero.

If there were a field in the conductor, then the charge in it would move as a consequence of that field. The charge would continue moving until the field was cancelled exactly.

Additionally, the field must be cancelled on the inner surface of the conductor, because if the charge cancels at some point inside the conductor, there is still an electric field between the surface of the cavity and that point, which means the charge between the surface and that point is going to move again until it cancels. Rinse and repeat until you hit the surface. Alternatively, a pretty simple energy argument has the same result: inside the conductor has a greater potential than the inner surface does, so the charge will naturally go to the lowest potential energy.

13. Mar 12, 2004

### Staff: Mentor

You are arguing that the field within the conductor is zero, but that's not what concerns me. (I take that as given!)

What I'm looking for is a simple argument showing that the field from the induced charge on the cavity surface exactly cancels the field from the charge within the cavity, independent of the charge distribution on the outer surface of the conductor.

14. Mar 12, 2004

Heh, I honestly wasn't sure what you were trying to get me to argue. It seemed to me I was just repeating myself!

Now that I see what you're really getting at, though, I think you're stuck with uniqueness theorems and Laplace's Equation. Once you find a solution, that's the only solution.

15. Mar 12, 2004

### Staff: Mentor

Yeah, I wondered if you missed my point or just thought I was dense.
Yep, that's all I've got. But it's pretty cool that it's true.

16. Mar 12, 2004

### paul11273

So...to answer the original question posted by frznfire219, does this mean that you can treat the problem as if the charge is placed at the center of the spherical conducter?

17. Mar 12, 2004

Yes.

18. Mar 12, 2004

### paul11273

So what's stated above will work to answer this question?

19. Mar 12, 2004