Electric field outside a spherical shell?

[solzonmars]

If a charge is put inside a spherical shell, why is the electric field outside the shell independent of the location of the charge? Gauss's law could find that the net flux is independent, but not each individual field?

Is this something about the surface charge density being independent of the location of the charge (if so, why does this hold)?

 

[kuruman]

Is this a conducting shell? If yes, then consider a Gaussian entirely between the inner and outer surface of the conductor. The electric field is zero everywhere on the surface therefore the electric flux through the surface is zero. By Gauss's law this means that the the total enclosed charge by the Gaussian surface is zero. Well, you know that there is charge Q inside so there must be total charge -Q on the inner surface to get a net of zero. Moving the inside charge Q around redistributes the charge on the inner surface but does not affect the electric field outside the shell. Stated differently, because the electric field inside the conductor is zero, changes in the electric field inside cannot be communicated to the electric field outside.

 

[solzonmars]

Yeah, it's a conducting shell.

I understand that the charge on the inner surface is -Q, which implies that the charge on the outer surface is Q. However, why would the fact that the electric field inside the conductor is 0 result in no effects on the outside—if the inner surface charge distribution changes then the outer surface charge distribution would change as well.

 

[Ibix]

However, why would the fact that the electric field inside the conductor is 0 result in no effects on the outside

Because there is no electric field from interior charge plus inner surface charge. It can't affect anything.

if the inner surface charge distribution changes then the outer surface charge distribution would change as well.

Sure. But why would the inner charge distribution change? Its distribution is driven by the enclosed charge's electric field to zero.

 

[PeroK]

Yeah, it's a conducting shell.

I understand that the charge on the inner surface is -Q, which implies that the charge on the outer surface is Q. However, why would the fact that the electric field inside the conductor is 0 result in no effects on the outside—if the inner surface charge distribution changes then the outer surface charge distribution would change as well.

I think it's a good question. The charge distribution on the inner surface is not uniform (unless the point charge is at the centre of the shell). But, the inner charge is distributed in such a way that the electric field inside the conductor is zero, as it must be. This means that the charge distribution on the outer surface must be uniform, as the outer surface is an equipotential.

But, there is a twist. Just because the electric field due to the point charge and the inner surface is zero inside the conductor doesn't immediately imply that the field is zero outside the shell (leaving only the influence of the symmetric outer charge distribution). One argument, however, is that you could imagine a very thick conducting shell, where the outer surface is a long way away. The inner surface charge in that case must be distributed so that the field is zero not just on a thin shell but for an arbitrary large volume. Then we can invoke a combination of the superposition principle and uniqueness theorems to say that the inner charge distribution must be such that its field combined with the field of the point charge is zero everywhere outside the inner surface.

That, then, leaves the uniform charge distribution on the outer surface as the only factor for the field outside the shell.

There is something rather deep and wonderful, IMO, about this whole thought experiment.

 

[Ibix]

But, there is a twist. Just because the electric field due to the point charge and the inner surface is zero inside the conductor doesn't immediately imply that the field is zero outside the shell (leaving only the influence of the symmetric outer charge distribution).

It follows immediately from Gauss' theorem, surely. Or am I missing your point?

 

[PeroK]

It follows immediately from Gauss' theorem, surely. Or am I missing your point?

Gauss's theorem shows that the total flux through a surface is zero, but you don't immediately have spherical symmetry to show that this means the field is zero everywhere.

Compare the case of a dipole.

 

[vanhees71]

I guess you mean a conducting shell (otherwise your statement is not correct). First let's get a qualitative picture, what's happening. If you bring the charge inside the shell the conduction electrons (assuming a usual metal as material) rearrange until they reach a static configuration, i.e., all the electrostatic forces on any conduction electron add up to 0, so that it doesn't move anymore (that's the condition for the field being static).

Now this example is one of the few examples which can be solved analytically and exactly, and it's well worth to study it carefully. Here some hints. As all electrostatics it starts with the potential and Poisson's equation
$$\Delta \Phi=-\frac{1}{\epsilon_0} \rho,$$
which has to be solved under the appropriate boundary conditions. Here the details depend on whether your shell is grounded or not.

Grounded shell: Then some conduction electrons can flow from or some electrons from outside can flow to the shell "to/from infinity", and you get the boundary condition
$$\Phi(\vec{x})|_{\vec{x} \in S}=0.$$
This makes the solution of the problem unique, and it can be found by the method of image charges. You'll find that there's a non-zero total surface charge on the shell (the total charge can be easily calculated using Gauß's law in integral form, the surface-charge distribution by calculating the jump of the normal component of the electric field across the shell).

Isolated shell: Then no charges can flow to or from the shell, i.e., the conduction electrons in the shell just rearrange. Here it's intuitively clear that you find the solution by using the solution for the grounded shell and just add the total influence charge sitting on the grounded shell as a homogeneous surface charge on the shell. This is because these charges don't feel a net force because the influence charges make the force balance while due to the homogeneous distribution and the corresponding symmetry the additional charges don't provide any net force on each other.

 

[kuruman]

Gauss's theorem shows that the total flux through a surface is zero, but you don't immediately have spherical symmetry to show that this means the field is zero everywhere.

Compare the case of a dipole.

I don't think so. For a closed Gaussian surface that lies entirely inside the conductor ##\int \vec E \cdot~\hat n~dS## is zero because ##\vec E=0## regardless of the position of the charge in the cavity. The conductor is an equipotential and ##\vec {\nabla}\varphi=0##. Given that the charge in the cavity is +Q and if the shell is initially neutral, there will be total charge -Q on the inner surface of the cavity and +Q on the spherical outer surface. By symmetry, the outer charge will be distributed uniformly over the surface. By Gauss's law, the electric field outside the surface is that of a point charge regardless of the position of the charge inside the cavity. That's because if you move the inside charge, the outside charge distribution cannot "know" that the charge has been moved because the only means of communication is the electric field and that is always zero inside the conductor. If dipole is placed inside the cavity, there will be a dipolar-like field inside depending on the dipole's position while the field outside will be zero according to Gauss's law.

 

[PeroK]

I don't think so. For a closed Gaussian surface that lies entirely inside the conductor ##\int \vec E \cdot~\hat n~dS## is zero because ##\vec E=0## regardless of the position of the charge in the cavity. The conductor is an equipotential and ##\vec {\nabla}\varphi=0##. Given that the charge in the cavity is +Q and if the shell is initially neutral, there will be total charge -Q on the inner surface of the cavity and +Q on the spherical outer surface. By symmetry, the outer charge will be distributed uniformly over the surface. By Gauss's law, the electric field outside the surface is that of a point charge regardless of the position of the charge inside the cavity. That's because if you move the inside charge, the outside charge distribution cannot "know" that the charge has been moved because the only means of communication is the electric field and that is always zero inside the conductor. If dipole is placed inside the cavity, there will be a dipolar-like field inside depending on the dipole's position while the field outside will be zero according to Gauss's law.

Well, that's another argument that the charge on the outer shell is uniform. And a good one.

But, fundamentally, the configuration as initially described does not have spherical symmetry, so you need some argument to impose spherical symmetry on the outer shell.

Your argument is that the inner configuration can only be communicated by the E-field and since it's zero inside the conductor, the outer shell must be uniformly charged.

My argument (although it's not my original idea!) is that you could move the outer shell arbitrarily far away and the field due to the inner charges must be zero over an arbitrarily large radius.

The field outside is uniform according to Gauss's law only once you have, by one argument or another, imposed spherical symmetry on the outer shell.

 

[PeroK]

For a closed Gaussian surface that lies entirely inside the conductor ##\int \vec E \cdot~\hat n~dS## is zero because ##\vec E=0## regardless of the position of the charge in the cavity. The conductor is an equipotential and ##\vec {\nabla}\varphi=0##. Given that the charge in the cavity is +Q and if the shell is initially neutral, there will be total charge -Q on the inner surface of the cavity and +Q on the spherical outer surface. By symmetry, the outer charge will be distributed uniformly over the surface. By Gauss's law, the electric field outside the surface is that of a point charge regardless of the position of the charge inside the cavity. That's because if you move the inside charge, the outside charge distribution cannot "know" that the charge has been moved because the only means of communication is the electric field and that is always zero inside the conductor. If dipole is placed inside the cavity, there will be a dipolar-like field inside depending on the dipole's position while the field outside will be zero according to Gauss's law.

So, here's my question:

The field inside the conductor is 0. But, how do you know that this is created only by a combination of the point charge and the inner shell? What if the zero field was created by a combination of the point charge, the inner shell and a non-uniform distribution on the outer shell? You might need all three fields to cancel out the field inside the conductor.

Why assume that the outer shell is not involved in zeroing the field inside the conductor?

 

[kuruman]

But, fundamentally, the configuration as initially described does not have spherical symmetry, so you need some argument to impose spherical symmetry on the outer shell.

The statement of the problem begins with,

If a charge is put inside a spherical shell ...

My argument (although it's not my original idea!) is that you could move the outer shell arbitrarily far away and the field due to the inner charges must be zero over an arbitrarily large radius.

What's wrong with that? The problem talks about a spherical shell which implies that the inner and outer radii are of the same order of magnitude. Even if you moved the outer radius to infinity, the field at infinity will be zero and all fields are zero at infinity.

The field outside is zero according to Gauss's law only once you have, by one argument or another, imposed spherical symmetry on the outer shell.

In what case? Certainly the field outside is not zero if there is charge +Q inside the cavity. If there is a dipole inside the cavity then yes.

 

[PeroK]

Certainly the field outside is not zero if there is charge +Q inside the cavity.

Yes, I meant symmetric, not zero.

Note that the problem is not spherically symmetrical. The point charge is asymmetrically positioned relative to the sphere. And, of course, in general the sphere could have an asymmetrical cavity. Only the outer surface needs to be spherical.

Let me try one last time to be understood.

a) You might claim that an E-field of zero could be created inside the conductor by a combination of the point charge and a (non-uniform) distribution on the inner surface.

b) I might claim that an E-field of zero could be created inside the conductor by a combination of the point charge, a (non-uniform) distribution of charge on the inner surface and a non-uniform distribution of charge on the outer surface.

So, you need an argument to show that a) has a solution. And, you need the uniqueness theorem to show that given a) has a solution, there can be no additional solution like b).

By the way, I think this is at the heart of the OP's question.

Note also, that this very point is discussed by Griffiths. To quote him: "perhaps nature prefers some complicated three-way cancellation".

Moving the outer surface away is the argument by which we see that a) has a solution. Then, you superimpose a uniform outer shell on your a) solution. The uniqueness theorem does the rest.

 

[kuruman]

So, here's my question:

The field inside the conductor is 0. But, how do you know that this is created only by a combination of the point charge and the inner shell? What if the zero field was created by a combination of the point charge, the inner shell and a non-uniform distribution on the outer shell? You might need all three fields to cancel out the field inside the conductor.

Why assume that the outer shell is not involved in zeroing the field inside the conductor?

Because of Gauss's law which takes the form ##0=\frac{q_{encl.}}{\epsilon_0}## for a Gaussian surface inside the shell. I already know that the Gaussian surface encloses charge +Q therefore to get zero net enclosed charge, there must be charge -Q on the inner surface. What's outside does not count as far as Gauss's law is concerned. As to why the outer distribution must be uniform, I need to think of a convincing argument for which I don't have the time right now.

 

[kuruman]

OK, time for the big guns. First the formulation of the problem. We have a spherical conducting shell of outer radius ##b## and inner radius ##a##. Some charge is placed inside the cavity at some distance away from the center. There are no charges in the region ##r>b##. Show that the surface charge density on the outer surface of the shell is uniform.

Solution
The conductor is an equipotential. Let ##\varphi_0## be its potential relative to infinity. Laplace's equation is valid outside the shell. If we choose the z-axis to be along the line joining the charge to the center of the shell, then the problem has azimuthal symmetry about that axis. This simplifies the formulation because we can use Legendre polynomials instead of spherical harmonics. The most general solution in the region ##r>b## is
$$\varphi(r,\theta)=\sum_{l=0}^\infty\left( \frac{A_l}{r^{~l+1}}+B_l ~r^l \right)P_l(\cos\theta).$$The boundary condition at infinity demands that ##B_l=0## for all values of ##l##. Then$$\varphi(r,\theta)=\sum_{l=0}^\infty \frac{A_l}{r^{~l+1}}P_l(\cos\theta).$$ We now apply the boundary condition at ##r=b##.
$$\varphi_0=\sum_{l=0}^\infty \frac{A_l}{b^{~l+1}}P_l(\cos\theta)=\frac{A_0}{b}+\frac{A_1}{b^2}\cos\theta+\dots$$For the potential to have the same value regardless of ##\theta##, the ##A_l## coefficients must vanish identically for ##l>0##. This gives ##A_0=\varphi_0b## and the potential in region ##r \geq b## is $$\varphi(r)=\frac{\varphi_0~b}{r}.$$This expression is a solution to Laplace's equation and satisfies the boundary conditions. By the uniqueness theorem, it is the solution.

The surface charge density is $$\sigma=-\epsilon_0 \left. \frac{\partial \varphi}{\partial r} \right|_{r=b}=\frac{\epsilon_0\varphi_0}{b}.$$ This charge distribution is uniform, Q.E.D.

 

[PeroK]

OK, time for the big guns.

If you prefer the big guns, I prefer the jiu jitsu.

Suppose we have a spherical conductor with a cavity cut out and a charge in the cavity. The charge distribution on the inner surface will be such that the field outside the inner surface due to the combination of the charge in the cavity and the inner surface is zero. And, the charge distribution on the outer surface will be uniform.

Proof

If we imagine a situation where the outer surface is a long way away (i.e. consider a very large sphere), then it has negligible effect on the field near the inner surface. And, by considering an arbitrary large sphere, we can see that the field outside the inner surface can be made zero everywhere.

There must, therefore, be a charge distribution on the inner surface that cancels out the charge in the cavity such that the field external to the cavity is zero everywhere.

Also, the field inside a uniformly charged sphere is zero.

So, we can superpose these two configurations to give a solution, hence the unique solution, to the problem.

QED

 

[kuruman]

Big guns or jiu jitsu, we agree; the surface charge distribution is uniform on the outside spherical surface.

 

[vanhees71]

Hm, the solution in #15 does not describe the solution for the given problem, which is most easily solved with the method of images. The charge distribution on the surface is only homogeneous, if the point charge in the interior of the shell is in the center of the sphere.

Of course, the solution is correct for the exterior of the shell, which is simply the Coulomb potential as if the charge in the interior of the shell were located at the center, but that's not the case for the solution inside the shell, and the surface-charge distribution is given by the jump of the normal component of ##\vec{D}=\vec{E}/\epsilon_0##, and this turns out not to be a uniform distribution, if the charge is not sitting in the center of the sphere. The complete solution of the problem of the insulated shell with 0 net charge is given here (but it's in German and using Heaviside-Lorentz units):

https://th.physik.uni-frankfurt.de/~hees/faq-pdf/estat-kugel.pdf

The surface-charge distribution is given in Eq. (13).

 

[PeroK]

Hm, the solution in #15 does not describe the solution for the given problem, which is most easily solved with the method of images. The charge distribution on the surface is only homogeneous, if the point charge in the interior of the shell is in the center of the sphere.

As I understand it, the OP considers an ungrounded, neutral shell.

 

[vanhees71]

Yes, and then #15 is not the complete solution for the problem. Particularly, if the pointcharge inside the shell is not in the center, the surface charge distribution on the shell is not uniform. It's given by the jump of the normal ocmponent of ##\vec{E}##, and the field inside the shell is not a Coulomb field around the center (see #18).

 

[PeroK]

Yes, and then #15 is not the complete solution for the problem. Particularly, if the pointcharge inside the shell is not in the center, the surface charge distribution on the shell is not uniform. It's given by the jump of the normal ocmponent of ##\vec{E}##, and the field inside the shell is not a Coulomb field around the center (see #18).

In that case, there is a non-uniform charge distribution on the inner surface of the shell and a uniform distribution on the outer surface.

 

[vanhees71]

As I understand it, the OP considers an ungrounded, neutral shell.

Yes, and that's precisely what's completely calculated in my manuscript. The surface charge is NOT uniform!

 

[vanhees71]

In that case, there is a non-uniform charge distribution on the inner surface of the shell and a uniform distribution on the outer surface.

Ok, if you have a shell of finite thickness, that's of course correct.

 

[kuruman]

Hm, the solution in #15 does not describe the solution for the given problem, which is most easily solved with the method of images. The charge distribution on the surface is only homogeneous, if the point charge in the interior of the shell is in the center of the sphere.

Post #15 was not meant to be a direct solution to OP's question. It was meant to show formally that the charge distribution on the outer shell is decoupled from the distribution on the inner shell, a point that we all agree on.

 

[PeroK]

Yes, and that's precisely what's completely calculated in my manuscript. The surface charge is NOT uniform!

I guess this is effectively an analytical proof of what we have all been assuming: that, wherever the charge is located, there exists a surface distribution that neutralises it.