Hollow conducting sphere in external E field

[AxiomOfChoice]

Suppose I put an (uncharged) hollow conducting sphere with inner radius a and outer radius b in an external electric field in the positive z direction.

Apparently, the potential INSIDE the sphere -- that is, for 0 < r < a -- is zero. Can someone explain why this is? I understand that the conductor is an equipotential, so we can set the potential for a < r < b to zero. But does this have any bearing on whether the potential is zero for 0 < r < a?

 

[AxiomOfChoice]

I have another question, related to this one. In Griffiths' E&M on pg. 141, he considers a closely related problem, except the sphere isn't hollow. He argues that, "by symmetry, since V = 0 on the sphere, the entire x-y plane is at zero potential." But in the NEXT SENTENCE, he says that the potential does not go to zero at infinity! WHAT?! If the potential is zero on the ENTIRE x-y plane, it's clearly zero as x,y approach infinity...right?

And I do not understand his symmetry argument...WHY is this the case? What symmetry is being appealed to here? (I've always hated symmetry arguments anyway...they always strike me as, "Well, I can tell just by looking at this that X has got to be the case, without calculating anything...can't you?")

 

[tiny-tim]

Hi AxiomOfChoice!

Suppose I put an (uncharged) hollow conducting sphere with inner radius a and outer radius b in an external electric field in the positive z direction.

Apparently, the potential INSIDE the sphere -- that is, for 0 < r < a -- is zero. Can someone explain why this is? I understand that the conductor is an equipotential, so we can set the potential for a < r < b to zero. But does this have any bearing on whether the potential is zero for 0 < r < a?

The internal free charge inside the uncharged conductor rearranges itself so as to cancel the field inside the conductor (a < r < b), not the potential.

And the field inside the sphere (0 < r < a) is zero.

I have another question, related to this one. In Griffiths' E&M on pg. 141, he considers a closely related problem, except the sphere isn't hollow. He argues that, "by symmetry, since V = 0 on the sphere, the entire x-y plane is at zero potential." But in the NEXT SENTENCE, he says that the potential does not go to zero at infinity! WHAT?! If the potential is zero on the ENTIRE x-y plane, it's clearly zero as x,y approach infinity...right?

And I do not understand his symmetry argument...WHY is this the case? What symmetry is being appealed to here? (I've always hated symmetry arguments anyway...they always strike me as, "Well, I can tell just by looking at this that X has got to be the case, without calculating anything...can't you?")

I don't have Griffiths, but I'm guessing that there's something about the problem that makes z = 0 a plane of symmetry, and that he means that the potential does not go to zero as z -> ±∞.

 

[AxiomOfChoice]

Thanks for trying to help me out, tiny-tim!

The internal free charge inside the uncharged conductor rearranges itself so as to cancel the field inside the conductor (a < r < b), not the potential.

And the field inside the sphere (0 < r < a) is zero.

I understand the FIELD inside a conductor is zero, but since conductors are equipotentials, can't we set the potential to be zero for a < r < b with impunity, since that's where the conductor is? If so, why does that tell us ANYTHING about the potential for 0 < r < a?

I also know the electric field has to be zero for 0 < r < a by Gauss's law. Hmmm...does THAT tell me V = 0 inside, since the potential has to be continuous, and the potential inside the sphere has to be a constant...which HAS to be zero, since the potential for a < r < b is zero?

I don't have Griffiths, but I'm guessing that there's something about the problem that makes z = 0 a plane of symmetry, and that he means that the potential does not go to zero as z -> ±∞.

Since the origin is the center of the sphere (which has radius R), the x-y plane *IS* a plane of symmetry...but so are (unless I'm very much mistaken) the x-z plane, and the y-z plane! Why isn't the potential zero everywhere along THOSE guys?

 

[tiny-tim]

Hi AxiomOfChoice!

Yes, you can set the potential V inside the conductor to be zero, but I certainly wouldn't do it "with impunity".

Hmmm...does THAT tell me V = 0 inside, since the potential has to be continuous, and the potential inside the sphere has to be a constant...which HAS to be zero, since the potential for a < r < b is zero?

Yes, V must be the same in the cavity as inside the conductor itself.

 

[TobyC]

There's a good explanation in the Feynman lectures about this.

If you consider a surface lying within the conducting sphere, say another sphere with a radius between a and b, then since the electric field is zero everywhere within the conductor, the flux of the electric field through this surface will be zero. By Gauss' law this means that the total charge enclosed by this surface is zero, and so the charge on the inside surface of the hollow sphere is zero.

Not only is the overall charge on the inside surface of the sphere zero, but there is no charge of any sign anywhere on the inside surface. If there were positive charges in some places and negative charges in other places then moving a charged test particle through the cavity from one location to the other would result in work being done, while moving it through the conductor to the same place would result in no work being done (since there are no fields in the conductor). This would violate conservation of energy.

The fact that there are no charges on the inside surface means that the situation of a conductor with a cavity is no different to a conductor without a cavity in terms of the fields produced inside. If you take a solid conducting sphere it will have no field inside and you can remove a spherical block from inside it without affecting the charge distribution, and therefore the field.

 

[AxiomOfChoice]

Not only is the overall charge on the inside surface of the sphere zero, but there is no charge of any sign anywhere on the inside surface. If there were positive charges in some places and negative charges in other places then moving a charged test particle through the cavity from one location to the other would result in work being done, while moving it through the conductor to the same place would result in no work being done (since there are no fields in the conductor). This would violate conservation of energy.

Thanks very much for your response, but this confuses me a little. Why can't the charge be arranged on the inner surface such as to exactly cancel out any electric field inside the cavity?

Also, I'm puzzled by the fact that in one case, you talk about moving a test charge from a point in the cavity to another point in the cavity, and then you mention moving the charge through the conductor to the same place. This seems nonsensical; how can I get to "the same place" without leaving the conductor?

 

[vanhees71]

As I see it, the most easy way is to simply solve the boundary-value problem. You start from Maxwell's Eqs. for electrostatics

\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=0.

The first Eq. tells you that

\vec{E}=-\vec{\nabla} \Phi,

and the boundary condition at the inner and outer sphere that the tangential components of the electric field are continuous. Inside the conductor the current must vanish and thus the electric field. Thus, it's immediately clear that there's no field inside the outer boundary, i.e.,

\vec{E}(\vec{x})=0 \quad \text{for} \quad r=|\vec{x}|&lt;b.

The remainder is to calculate the potential for r&gt;b.

At infinitity you have

\vec{E} \simeq E_{\infty} \vec{e}_z \quad \text{for} \quad r \rightarrow \infty.

In terms of the potential, we have

\Phi \simeq -E_{\infty} z=-E r \cos \vartheta \quad \text{for} \quad r \rightarrow \infty.

At r=b the potential must be constant to make the electric field perpendicular to the boundary. We can choose this value to be 0.

The whole problem is rotational invariant wrt. to the z-Axis, i.e., the potential is independent of the azimuthal angle, \varphi, and thus, expanding in terms of electrostatic multipoles, the potential must take the form

\Phi(\vec{x})=-E_{\infty} r \cos \vartheta + \sum_{l=0}^{\infty} a_l \mathrm{Y}_l^0(\vartheta) r^{-(l+1)}.

For r=b this leads to

\sum_{l=0}^{\infty} a_l \mathrm{Y}_l^0(\vartheta) b^{-(l+1)}=E_{\infty} b \cos \vartheta.

From the orthogonality of the spherical harmonics, only a_1 \neq 0, and its value must be

a_1=2 \pi E_{\infty} b^3 \int_0^{\pi} \mathrm{d} \vartheta Y_l^{0}(\vartheta) \cos \vartheta=2 E_{\infty} b^3 \sqrt{\pi/3}.

Thus we finally get

\Phi(r,\vartheta)=E_{\infty} \frac{b^3-r^3}{r^2} \cos \vartheta=E_{\infty} z \left(\frac{b^3}{r^3}-1 \right).

That means that outside the sphere, r&gt;b, the induced field is that of a dipole in the z-direction. The total field is, of course, the superposition of the imposed homogeneous field and the induced dipole field.

 

[TobyC]

Thanks very much for your response, but this confuses me a little. Why can't the charge be arranged on the inner surface such as to exactly cancel out any electric field inside the cavity?

Also, I'm puzzled by the fact that in one case, you talk about moving a test charge from a point in the cavity to another point in the cavity, and then you mention moving the charge through the conductor to the same place. This seems nonsensical; how can I get to "the same place" without leaving the conductor?

Thanks very much for your response, but this confuses me a little. Why can't the charge be arranged on the inner surface such as to exactly cancel out any electric field inside the cavity?

Also, I'm puzzled by the fact that in one case, you talk about moving a test charge from a point in the cavity to another point in the cavity, and then you mention moving the charge through the conductor to the same place. This seems nonsensical; how can I get to "the same place" without leaving the conductor?

You're absolutely right, the argument I posted was flawed. It's not my argument anyway, it's from the Feynman lectures, but I was misinterpreting it.

Here is the modified (I hope correct) version of my original post:

If you consider a surface lying within the conducting sphere, say another sphere with a radius between a and b, then since the electric field is zero everywhere within the conductor, the flux of the electric field through this surface will be zero. By Gauss' law this means that the total charge enclosed by this surface is zero, and so the overall charge on the inside surface of the hollow sphere is zero.

If there are no charges of any sign on the inside surface, then there is no field in the cavity since the situation is no different to having a solid conductor. If there are positive charges in some places and negative charges in others then there can still be no no field in the cavity because the field lines through the cavity would have to extend from the positive to the negative charges and so work would be done in going through the cavity, while no work would be done if you went to the same point on the surface through the conductor.

There is therefore no field within the cavity. (There are also no charges on the inside surface).

The best way to prove this result though is I think to use vector calculus like the poster above. If there are no charges in the cavity then the potential within the cavity satisfies Laplace's equation and it is a constant on the boundary of the cavity (since the boundary is a conductor). This has a unique solution which is the potential being constant throughout the cavity (no electric field).