Expert: Daniel Mazur - 10/13/2009

Question
QUESTION: Hi Daniel,

There are two conceptual problems that have been worrying me since pretty long. I've tried to reason them out and tried to figure it out many times but now I'm losing patience!!

The first problem that I have is related to Capacitors:

Suppose we have a parallel plate capacitor fully charged and connected to a battery. Let the plates have a charge Q and be seperated by a distance d.

Now when we move the capacitor plates further apart and increase the distance between them from d to d', some of the charge from the capacitor plates flows back to the batteries.
The common explanation for it is that since the capacitor is connected to a battery which has a fixed pot. diff., it has to maintain that pot. diff. and that's what it does when it reduces the charge on itself when the plates are moved apart.

This a nice way of stating it. But I don't understand what goes on. Electrons surely don't understand the above explanation. Nor do they understand conservation of energy. But they obey the coulomb's law. So they flow only when there is a net electric force on them. In this case, I don't see where that force comes from.
The plates are considered to be infinite so they have a constant field equal to rho/2e. So doesn't matter even if we pull the plates apart the field still is the same.
So basically I don't understand, that what is it that makes the electrons from the negative plate to move towards the battery which in turn causes the positive charge from the positive plate to move towards battery.

Question 2:

This is something similar to the above. I don't understand what causes it.
Suppose we have three concentric thin metallic spherical shells.
They have a radius say a,b and c where c>b>a.

Also the middle shell i.e. the one with the radius b, is grounded.
The smallest and the biggest shells(i.e. with radius a and c resp.) have a charge
+q and -q , respectively on them.
Now it has been asked to find out the final charge distribution on all the surfaces of the shells.
If we apply gauss's law on the middle shell, the field inside it should be 0 as it is a conductor, so a charge -q appears on the inner surface of this middle shell which along with the +q charge on the smallest and the innermost shell makes the net charge enclosed by the middle shell 0.
There is a solution given to this problem in which the key point that has been used is that since the middle shell is grounded, the potential at it will be 0 due to all other charge distributions. And working with this idea we get a charge distribution
on the leftover surfaces as follows:

On the outer surface of the middle shell : +bq/c

On the inner surface of the outermost shell: -bq/c

On the outer surface of the
outermost shell: -q(1 - b/c)

Now again what I don't understand is that how did the charges appear on these leftover surfaces given above?
Why can't the distribution stop with just the inner surface of middle sphere getting a negative q to balance the positive q of the smallest shell.
Yes, the potential won't be 0 at the middle sphere if the distribution were to stop here.

But there is no electric field present to cause a charge to appear on the outer surface of the middle shell because even though there is a charge -q present on the outermost shell, it's field inside is 0 as it is the case for any spherical shell. And so it's field on the middle shell lying within it is 0.

The charges present within this middle sphere are -q on it's inner surface and +q on the innermost shell. The fields of these charges will be mathematically same to that due to a single charge kept at the center. So there fields also cancel out since both have charges with opposite signs.

So where does the field come from???
I know that the potential at any point inside the largest shell is
-kq/c and the middle shell is also placed within this largest shell and so it is at this potential and at the same time is grounded. But then again, the electrons don't understand this!
They need a net field!
Please tell me where is it??

Thanks a lot,
Shikhin

ANSWER: Hi Shikhin,

the answer to the first question is relatively easy. But one must first understand, that the battery creates the potential difference by taking some conduction electrons from the wire going to one electrode and giving it to the wire leading to the other electrode. So, even if the electrodes are at an infinite separation some charge difference is there, but it is equilibrating so that the free-charge densities are the same in the electrodes as in the wires connecting them to the battery. Naturally, if the capacitor plates are infinitely large the wires will contain comparatively negligible amount of the total charge.

As we bring the electrodes closer, the charge carriers will feel the presence of the electrostatic field created by the other electrode and so they try to pack at a higher density inside the electrodes. The electrons (in the negative electrode), however, repel each other, so that there is always such an equilibrium, where the external (from the other electrode) Coulomb field is compensated by the internal (Coulomb repulsion between electrons) field. You see, there is a balance of forces at every distance between the electrodes, although each balance corresponds to a different charge density (or total charge, if you wish) in the electrodes. In the positive electrode the effective positive charge acts essentially like a mirror image of the electrons, even though it is a bit more abstract effect.

To fit the reasoning with some equations, the two plates of the capacitor are at a constant potential difference U = phi2-phi1, because the battery makes it so. At a separation d the electrostatic field E between the charges in the electrodes is E = U/d. The force with which the charges (assuming elementary charge carriers) are pulled together is F = e*E and this force can be measured, because it becomes the mechanical force with which the electrodes attract each other (and with which we must hold the electrodes to stay apart). This is the solution: With changing distance from d to d' we change the external force from F to F' = e*E' = e*U/d' . If d'>d, the external force becomes smaller and some charge will flow away pushed by the internal forces until a new equilibrium is reached.

The second question took me some time to think through. The grounding of the middle sphere is making that the potential there is fixed at ground (it is tricky here to call it zero, because it will not be, strictly speaking!) potential, and at the same time it allows the total charge on that sphere to be non-zero - as induced by the other two spheres. The problem speaks of thin spheres, and that presumably means just that we neglect the difference between the radii of the inner and outer surfaces of the spheres.
I take it that you have no problem with the inner surface of the middle sphere being charged +q, your questions is: Why the outer surface charges at all? Why doesn't it stay neutral, when everything at r<b compensates and it doesn't (following Gauss) feel the charge at the outer shell?

The reason why the above suggestion cannot be the solution is, that not only charges (in their absolute amounts) but also potentials generated by the charges need to equilibrate. This is quite logical, the electrostatic potential of the outer sphere (below k=1/(4*pi*eps0)) is +kq/c. The potential of the inner sphere (-q) and middle sphere (+q) together would be 0, if additional charge didn't flow. This potential difference could do nothing, if the middle sphere was isolated (and presumably charged +q), but as it is connected to ground, which acts as a reservoir of charge, the middle sphere must take on some additional charge in order to raise its potential so much that the resulting VOLTAGE (potential difference) between itself and the outer shell becomes zero.

You could imagine the outer sphere as an immaterial shell at a potential +kq/c. Then the middle+inner shells need to take on so much net charge that if b expands all the way to b=c, the middle shell will be at the same potential. I presume that you know the equations, but just to make the reasoning complete:
U_bc = kq_b/b - kq_c/c = k(q_b/b - (+q)/c) = 0 is the principal requirement.
From this follows that q_b = +q(b/c). The q_b here is, naturally, the net charge of the middle+inner sphere as seen by the outer sphere.

Once this calculation is complete, the charge distribution on the surfaces of the outer shell are straightforward: The inner surface must carry -q(b/c) to compensate for the field of the net charge inside it and the outer surface must carry +q+q(b/c)... In this number you and I differ, but I don't see, how the outer surface can have -q(1-b/c) if the net charge of the outer sphere (i.e. its inner plus its outer surface) is to add up to +q. Please check it.

All best,
Daniel


---------- FOLLOW-UP ----------

QUESTION: Hi Daniel,

Thanks so much for the reply.
But I still have a few points unclear.

Question1:

[Q] As we bring the electrodes closer, the charge carriers will feel the presence of the electrostatic field created by the other electrode and so they try to pack at a higher density inside the electrodes.  [/Q]

What is unclear to me here is that how can the electrons feel a force on briging the plates closer.
Why I say this is because the field due to an "infinite" plate is given by
E = rho/2(eps_0)

where rho stands for the surface charge density.
This field is same at every point in space and doesn't increase or decrease if we move away or towards it.
So even if we move one plate closer or away from the other, the force due to this plate on the charges inside the other plate remains the same.
So how do the charges in one plate experience a force due to the other plate as we move the two together, as you stated.

Question2:

[Q]
the middle sphere must take on some additional charge in order to raise its potential so much that the resulting VOLTAGE (potential difference) between itself and the outer shell becomes zero.
[/Q]

But there is no potential difference between the middle sphere and the outer sphere.
For thin spherical shells having uniform charge distribution, the electric field at any point inside them due to their surface charge distribution is 0.

So the net electric field "inside" the outermost shell is always 0 no matter what charge we give to the outermost shell.
The two other smaller shells also lie completely "inside" the outermost shell and so they have 0 electric force on them due to the charge on the outermost shell.
So there is no potential difference between them.

Also why your charge distribution differs from that I wrote is because you took the charge on the outermost sphere to be +q, perhaps accidently, whereas in the problem it is given to be -q.
That's why there's a difference of + and - in your and my result.


So basically there is no "potential difference" between the middle and the outermost shell.
The potential at the middle shell due to the -q on the outermost shell will be the same as it will be on the surface of the outermost shell itself, i.e. -kq/c

So yes, though the middle shell is grounded and so is at a potential 0 and at the same time it's surface is at a potential -kq/c due to the charge on the outer shell, and so to make its potential at its surface equal to 0, it draws some charge from the ground, but I cannot see any "net electric field" acting on the middle shell which makes it pull some charge up from the ground.
As I already wrote, that the electric field due to the -q charge on the outer shell is completely 0 inside it and so it can't be the source of the field.
Also the field due to the +q on the innermost shell is exactly cancelled out by the -q residing on the "inner surface" of the middle shell as both these charge distributions are contained inside the middle shell and thus their field is mathematically equal to that due to a point charge placed at the center.

So finally considering all possible fields "on the middle sphere":

Due to -q on the outermost shell : 0

Due to +q on the innermost shell: +kq/(b^2), where b is
         the radius of the
         middle shell.

Due to -q residing on the "inner surface" of the middle shell: -kq/b^2

So the "fields" due to all the charge distributions sum up to 0 "at the middle shell". But still the middle shell draws up some charge from the ground!
Working it out by considering the net potential due to all charge distributions at the middle sphere is one way....which has also been given in the solution in my book....and which you also used in your reply....and that's absoluetly correct and okay but then at the same time for the charges to get pulled up from the earth, we need a net electric field that acts on them..!!
And there is 0 net electric field acting on the middle sphere as I showed above....and still the charges get pulled up!!
How??

Thanks,
Shikhin

Answer
Hi Shikhin,

Question 1:
[Q]
This field is same at every point in space and doesn't increase or decrease if we move away or towards it. So even if we move one plate closer or away from the other, the force due to this plate on the charges inside the other plate remains the same. So how do the charges in one plate experience a force due to the other plate as we move the two together, as you stated.[/Q]
The force is F = eE = e*grad(Phi) and the potential difference (voltage) between the plates is constant U=Phi2-Phi1. The "grad(Phi)" for a plate capacitor is (Phi2-Phi1)/D, where D is the separation distance between the plates. So, if you increase or decrease the separation D, the grad(Phi), E and F all decrease or increase, respectively. You are trying to reason using constant separation arguments in a problem with variable separation. That cannot work.

Question 2:
[Q]
But there is no potential difference between the middle sphere and the outer sphere. [/Q]
Yes, there is. The elstat potential of a spherically symmatrical distribution of charges Q lying at radii smaller than or equal to R is defined as Phi(R) = k*Q/R, we cannot get around that. This defines the elstat potential of the outer sphere at k*(+q)/c. Supposing the inner sphere has (-q) and the middle sphere only carries the induced (+q), then the total charge enclosed by a sphere of R=b is 0 and therefore Phi(b)[inner=-q,middle=+q]= k*(-q+q)/b = 0. The k*(+q)/c above isn't zero in this problem, so there IS a potential difference U[inner=-q,middle=+q] = Phi(b)[inner=-q,middle=+q] - Phi(c) = 0-k*(+q)/c.

[Q]
For thin spherical shells having uniform charge distribution, the electric field at any point inside them due to their surface charge distribution is 0.
So the net electric field "inside" the outermost shell is always 0 no matter what charge we give to the outermost shell.
The two other smaller shells also lie completely "inside" the outermost shell and so they have 0 electric force on them due to the charge on the outermost shell.
So there is no potential difference between them.[/Q]
You keep considering only the total amounts of charges and dismissing the definition of electrostatic potential. Unless you accept that Phi(R) = k*Q/R (Q being net charge enclosed by shell of radius R), I am afraid that you cannot make any progress in understanding this problem.

[Q]...and that's absoluetly correct and okay but then at the same time for the charges to get pulled up from the earth, we need a net electric field that acts on them..!!
And there is 0 net electric field acting on the middle sphere as I showed above....and still the charges get pulled up!![/Q]
The problem of putting a test charge inside a charged shell and observing that there is no force acting on it is DIFFERENT from the question of force acting on a charge OUTSIDE the charged shell (in the ground reservoir) to be dragged from outside in!! These are (like the confusion in Question 1) two different problems and have different solutions. Perhaps here lies the key to overcome your roadblock.

I'll let you give it some more thought for the moment... :-)

Good luck!
Daniel