Expert: Daniel Mazur - 10/20/2009

Question
QUESTION: Hi Daniel,

First of all I'm sorry for using nonsensical terminology.
I do understand, with absolute clarity, the difference b/w "potential at a point" and the "potential difference between two points"
in an electric field.
This is not the first time I've used nonsensical terminology. This happens with me half the time. In the process of thinking and typing I made some really stupid errors.

I also know that dV = - E.dr
But my writing V=E/d was nothing more than a typing error, or simply an error, that I didn't pay attention to.
I'm sorry for that. I will make sure I don't do such a thing in future.
I have done the topic of electromagnetism in quite some depth. Please believe me. That's just not the problem. I might be absolutely imprecise when I write things or try to communicate something but that's just because of a continous thought process going alongside while you are typing.

This is one last time I want to ask you the question but now with a different approach. If you still feel after this question that I really need an even deeper study of the topic, I'll surely do that and will come back to these problems some time later.
I want to approach the problem a little differently now.
First of all let me just state what you said again.
Please tell me if I'm anywhere wrong in it.

the reason for decrease in charge on movig the plates is because of the following :

The plates are at a constant "potential difference" , say V . This difference in potential is maintained because the plates are connected to the battery which itself has a constant pot. diff.

So because of the battery, the pot. diff. b/w the capacitor plates remains same.

Also E.d = V
or E = V/d
Now if d changes, E will change accordingly since V is constant.
And so obviously the charge reduces because that's why or that's how E reduces.

Am I wrong again here??


Ok, now I have a different situation. If you could give me an answer to this, I think it would solve most of my problems.

Suppose we have some positive charge distribution that produces a "constant" electric field, say E in the horizontal direction, lets say, towards the negative x-axis.
So the direction of this constant E is towards the -ve x axis and the field just lies in the horizontal direction.

Now we have two plates, just like we have in a parallel plate capacitor. But they don't face each other and it's not a capacitor like arrangement.
Now let one plate lie above the X - axis, i.e. having some +y coordinate and let the other one lie below the x axis.
So one plate has a +y coordinate and the other has a -ve y coordinate but at the same time let them be placed just directly above or below each other , i.e. both have the same x coordinate. Lets call this initial x coordinate of the plates A.
Also their surface areas are perpendicular to the field direction. They have two faces : the front and the back.
The area vector of the front surface is opposite to the field direction or in other words is towards the +ve x - axis.
So obviously the back faces have their area vectors pointing in the direc. of the field or the negative x direction.

Now let us connect the back faces of the two plates with a conductig wire and also lets place equal charges equal to -Q on both the plates seperately.

Now the -Q charges will try to move in a direction opp. to the fiedl direc. or in other words towards the +ve x direc.

So the -Q charges get concentrated on the front faces of both the plates.

Now let us move the upper plate, i.e. the one lying above the x-axis towards the left or towards the negative x direc.
Let this new x coordinate of the upper plate be B whereas the lower plate,i.e. the one lyig below the x axis, is still at x coordinate A.

Now what will happen is this: (I am not sure wether it will happen or not, do tell me if it doesn't, I'm just assuming that following is what happens and that's what my question is based upon.)

The charge from the upper plate moves through the conducting wire all the way upto the lower plate at x coordinate A.
I don't know wether the entire -Q will move or not as you talked about electron-electron replusions.
But let's say some of it does and then an equilibrium is achieved.
How I interpret it is this:
The charges have a net coulomb's attractive force on them due to the +ve charge distribution producing the field.
Now when the plates are at x coordinate A, what the charges can do at the most is get as near the +ve charge distribution as possible or in other words get concentrated at the front surface of the plates. They cannot break through the plate and move towards the +ve charge distribution.
But when the upper plate is taken a little backwards i.e. to the x coordinate B, the charges resinding on its surface get another path to move forward towards the +ve charge distribution, i.e. by moving through the conducting wire all the way up to A.
First of all am I right in the above or not?

This may be said in another way by sayig that the charges always try to decrease their pot. energy or since there is a pot. diff. between the two pos. A and B and since the plates are connected , the charges move from higher to lower potential.

What i don't undestand is this:
First let's consider the situation when both plates are at A.
Half the conduction electros of the wire lying in the lower part of the wire will try to get concentrated on the lower plate. same will happen with the upper plate.
So half the elctrons go on one side, and the other half on the other. And ofcourse all of them don't reside on the surface ...there are interal repulsions so an equilibrium state is achieved, but essentially the positioning of the elctrons in the upper as well as lower half is symmetrical.
Now when we move the upper plate to the position B the electrons start flowing down towards A from there.
Even though they still experience the same force they still flow.
Is it because when we move the upper plate to B , the shape of the wire changes and the vertical part or the curved part of the wire now lies horizontally and the electrons lying in this region that were earlier concentrated more near the upper plate, now move towards the lower plate, which in turn causes a partial positive charge in that reagion of the wire which in turn leads to the electrons concentrated near the upper plate to come down and neutralise the positive charge created in that region of the wire?
Is it then just by virtue of the shape that the wire takes that the charges flow towards the lower plate??

I know something similar happens in a capacitor also.
But I don't think it should be just becasue of the shape that wire takes when one plate is moved backwards.
It could be any shape. But then the electros don't understand that by moving in such and such direction they can reduce their potetial energy.

What I basically want to know is that how this charge flow take place....what triggers it... the sequence of events...

If you still think that I need to first get into an even more detailed study of electromagnetism..before we can discuss anything further on it....i will do that....

At the same time, it would be great if you could answer this question.

Thanks a lot,
Shikhin

ANSWER: Hi Shikhin,

I accept your explanation. At the same time I must require for you to not write vaguely and carelessly. We are talking about part of nature/science that is very well known and precisely described. I cannot imagine that you'd achieve anything but a fail, if you turned in a homework or a test written in the same manner in which you wrote to me. This is not pedantry, this is essential for me to understand, what you actually are missing and need to learn. It costs us both a lot of time in the best situation and I beg your common sense to stop sending it down the drain. I will answer to this topic once more, and please try to learn from the admonitions I will leave in instead of getting upset. It's all meant to show you the way.

Now to your recent writing:
[Q]And so obviously the charge reduces because that's why or that's how E reduces. Am I wrong again here??[/Q]
You are not wrong you are vague. So much so that the sentence carries no information about the reason for charge removal. There are equations to use, or writing the equations in words. Geometry of the system enters the equations via the capacity C = eps0*A/d . As you never used the key word "capacity" in your reasoning, I took it for not understanding the basics. The meaningful sentence would then be:
"And the charge reduces for increasing d, because q = U*C = U*eps0*A/d, where on the right-hand side everything except d is (or is kept) constant." Can you feel the difference?

I drew a picture of the new problem, thank you for describing it thoroughly as you did. Indeed, after you move the upper plate to x=B (where B<A), some negative charge will traverse to the lower plate.
[Q]...This may be said in another way by sayig that the charges always try to decrease their pot. energy or since there is a pot. diff. between the two pos. A and B and since the plates are connected , the charges move from higher to lower potential.
[/Q]
Your reasoning is also right in both views, while the one using potentials is better, cleaner, more rigorous.

[Q]What i don't undestand is this:
...
Now when we move the upper plate to the position B the electrons start flowing down towards A from there.
Even though they still experience the same force they still flow.[/Q]
The last sentence is wrong AND it contradicts the paragraph I quoted above - the one about potential difference. Where there is a potential difference, there is a net force. NEW net force, because the ORIGINAL equilibrium has a zero net force (= zero pot. difference between the plates), while the NEW situation before (!) new equilibrium is reached has a non-zero net force due to non-zero pot. difference.

[Q]Is it then just by virtue of the shape that the wire takes that the charges flow towards the lower plate??

I know something similar happens in a capacitor also.
But I don't think it should be just becasue of the shape that wire takes when one plate is moved backwards.
It could be any shape. But then the electros don't understand that by moving in such and such direction they can reduce their potetial energy.[/Q]
The shape of the wire plays no role in the essentials we are discussing now, this perception of yours is correct. Your problem with understanding the matter arises from the fact that even though you see the potential difference, you don't see the force (see above). I really don't know, how to remove this roadblock from your mind... Let's take an analogy.

Imagine two identical containers of water instead of charged plates and a tube instead of the wire. As long as the containers are at equal height in a gravitational field, they will hold the same amount of water. As you change the height (=change the gravitational potential) of one container, water starts flowing in order to minimize the energy of the entire system. New equilibrium is reached, when the water level in one container is again even with water level in the second one, because this arrangement means no more potential difference between the water bodies. In this problem also the shape of the connecting tube is immaterial.

[Q]But then the electros don't understand that by moving in such and such direction they can reduce their potetial energy.[/Q]
Electrons, just like water, don't NEED to understand anything. They naturally concentrate in the minimal energy arrangement within their container. When you create a path for them that would allow them to lower their energy, should they enter it, they will enter. Just like water... Electrons too have their repulsive forces, the environment inside the wire works in the sense of cohesive forces you see between molecules of water. The analogy goes further than you might think.

Another example. When you fall on your mouth, it's not because you understand that you are supposed to do so. It's just that your body obeys the law of lowering its potential (potential energy) in the grav. field. If this happens to you in your room, you just end up lying on the floor and nothing much is observed from the outside - you are still in the room. If it happens to you on a balcony, though, and you don't use your brain to avoid the edge, your body will - in spite of your intelligence - gladly minimize its energy through a free fall with drastic results for you. So, precisely because electrons have NO knowledge and can NOT think for themselves, they behave the way we see them behave.

[Q]What I basically want to know is that how this charge flow take place....what triggers it... the sequence of events... [/Q]
I think we have gone though this sequence well enough with the previous problems, but we can do it once more. You got the sequence correct until the point (noted above) of seeing the potential difference but not seeing the force. For the sake of clarity and simplicity, I will slightly modify your problem without changing any pertinent equation:

Let's start with conductive plates, one at a position A and the other at a position B<A, which are electrically isolated from one another. Each of them carries charge -Q, which is internally redistributed to account for the external field -E. We ask what happens the moment we suddenly connect the two plates with a wire.

Now YOUR first task is to realize that this IS a situation equivalent to the one you described. After you concede that, you may continue reading on and considering the following.

If a wire across an electrical field lies between points of different potential, the potential difference will distribute along its length. To electrons this presents a flight of stairs or a slide, a toboggan, down which they can tumble. And because they have no brains to stop them, they WILL tumble down the slide.

Now, as the electrons are both SUBJECT to an electrical potential (and field) and also SOURCES of electrical field, their increased presence in the lower-potential electrode starts lowering the slope of the toboggan. The "slope" corresponds to the potential gradient along the wire. Eventually, the equilibrium is reached: The externally applied potential difference becomes compensated by the internal redistribution of charges. Both plates and the wire connecting them will be on ONE and the same potential! You see, in the new equilibrium the two plates and wire together represent only ONE body, albeit with a complicated shape. The internal redistribution of charges is then driven by the same principles as a redistribution in a single conducting sphere (for example) in an external field. If you understand the reasons for charge redistribution in such a sphere, really understand, then no system of many conductively connected bodies in an external el. field can disturb you.

I do not think that you need any more electromagnetism as a whole. I think you need just more of the electrostatic part. It is not an entirely simple subject, because it considers materials. In materials, questions can easily get out of hand. But if you just work with the ordinary conductors (Ohm's law, free charges) and insulators (fixed charges), you will proceed. Do you know, what field distribution is in a simple wire, when you short the battery contacts with it? Do you fully understand the role of capacity? Can you see a net force every time there is a net potential difference? Do you know, what the condition "electrode ... is grounded" means for the solution of a problem? Etc. I do not underestimate it, these simple things can prepare you some very embarrassing moments in later professional life, if not grasped in time.

So, I don't know, if I helped at all, but again I think I did my best.

Good luck,
Daniel



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

QUESTION: Hi Daniel,

Thanks so much for the reply. I think i understood it.
I just want to confirm wether I've got it right or not.

Let's take a pair of like charges and another pair of unlike charges.
Now as we know by coulomb's law, likes repel and unlikes attract.
Let's throw a like charge towards the other. As the charge moves near the other, the KE, that we gave to it starts decreasing and we can't observe this decrease in energy reappearing somewhere else. But this energy doesn't vanish totally too, it comes back. So we define the pot. energy corresponding to the configuration of charges.
So if any of the pairs, like or unlike, is left to itself, it will start moving away or towards from each other, as the case may be.
So in any case, the system decreases it's pot. energy.
So the statement that the system will try to decrease it's pot. energy is just another way of saying that it will follow its natural tendency of attraction or repulsion.
Am I right when I say the above?'?
And if that's right then that is what happens in the situation that I gave.
When the two plates are at the same x coordinate, A, the electrons residing on the very outer surface of the plates want to move towards the positive charge distribution but they can't get out of the material. And the electrons lying well within the plate or the wire, they also experience the force due to the field towards the +ve x but they also experience an equal and opposite force from the elctrons collected near the surface.
Now whe we move the plate having a + y coordinate back to B, this equilibrium is temporarily broken.
And for how the equilibrium is exactly broken, I have the following picture. please tell me if it's correct.
Consider the case when both plates are at x=A.
Also consider the connecting wire to be a U shaped one, i.e. it starts from the upper plate, goes horizontally to some distance in the negative x direc. curves around, comes back horizontally to x=A and touches the lower plate.
So it's something like a stretched letter 'C' .
Now consider some elctrons lying in the upper horizontal part of the wire, i.e. near the upper plate. They experience a force due to the field towards the +ve x and an equal repulsive force due to the electrons collected near the upper plate in the -ve x direction.
Now when we move the upper plate to B, this horizontal part that initially lied above the x axis, now lies below the x axis(not necessary, but some part of the wire will move downwards, and I'm assuming it's that part) and these elctrons now only experience the force due to field in the +ve x direc. The -ve charge on the lower plate is not enough initially to provide an equal repulsive force. So the electrons start moving towards the lower plate until enough charge collects at the lower plate to provide a repulsive force, equal to the force due to the field, to the oncoming electrons. And this is where the new equilibrium is established.
And now we can of course put it in terms of potential.      

Am i right in the above or not.


Also I wanted to discuss the problem of the spheres.
As you said, the potential at the surface of a sphere of a radius R is kQ/R where, Q is the net charge enclosed by the surface of radius R.
But what if we also have a charge distribution at a radius R' where R'>R??
Isn't the potential at the sphere of radius R due to the charge q on the sphere of radius R' equal to kq/R'??
And if that's true, the potential at sphere of radius R is there due to both the charge distributions, i.e. due to the charges lying within it + the pot. due to the shperical charge distribution lying outside it.
If this is true, we can apply this to the three shells of our problem.
Consider the middle shell. It's inner surface has  -q and the inner shell has +q. So the net charge inside the middle shell is 0 and thus the potential at the surface of this middle shell , which has the radius b, is also 0 "due to the net charge inside". The outer shell, which has a radius c (c>b) which encloses the middle shell has a charge -q.
The potential due to this -q on the middle sphere is -kq/c.
So the total potential of the middle sphere is :
0 + (-kq)/c = -kq/c.
Also the net charge enclosed by the outer shell is also 0.
So the potential at the outer shell is olny due to the -q residing on its own surface which is -kq/c.

So we see that both the middle shell and the outer shell have the same pot. which is equal to -kq/c.

How do they have a pot. diff.??

Thanks a lot,
Shikhin

Answer
Hi Shikhin,

[Q]So the statement that the system will try to decrease it's pot. energy is just another way of saying that it will follow its natural tendency of attraction or repulsion.
Am I right when I say the above?'?[/Q]
Yes, the two statements are saying the same. You could have also said that there are forces in the system, which are not balanced (blocked) and so the system will evolve under the influence of these forces.

[Q]Now consider some elctrons lying in the upper horizontal part of the wire, i.e. near the upper plate. They experience a force due to the field towards the +ve x and an equal repulsive force due to the electrons collected near the upper plate in the -ve x direction.
Now when we move the upper plate to B, this horizontal part that initially lied above the x axis, now lies below the x axis(not necessary, but some part of the wire will move downwards, and I'm assuming it's that part) and these elctrons now only experience the force due to field in the +ve x direc. The -ve charge on the lower plate is not enough initially to provide an equal repulsive force. So the electrons start moving towards the lower plate until enough charge collects at the lower plate to provide a repulsive force, equal to the force due to the field, to the oncoming electrons. And this is where the new equilibrium is established.[/Q]
I am not sure you have the correct picture, because your interpretation seems to rely on the C-shaped connecting wire. It is not only the charge in the wire, which will get below the x-axis, that will flow into the lower electrode. Also some charge from the upper electrode will do so. The wire represents something like a duct containing an electron fluid. Even the electrons in the upper plate, who have to flow "up a potential hill" until they reach the bend of the C-shape, will gladly do so, because the net forces on them (total of internal and external) become suddenly unbalanced. As I have told you, you can use a water hose as and example - it's a good analogy of the behavior of electrons wanting to lower their energy. If the water flowing up a hill ina part of the hose does not disturb your conscience, the the electrons shouldn't either. But maybe you understand it already and only I am not convinced and deterred by your special choice of the shape of the wire.

[Q]But what if we also have a charge distribution at a radius R' where R'>R??
Isn't the potential at the sphere of radius R due to the charge q on the sphere of radius R' equal to kq/R'??[/Q]
No. This is one of the textbook examples of electrostatics and it is one of the simplest consequences of the Gauss law. First take the homogeneously charged spherical shell of radius R' alone with just empty inside. The potential all across the inside will be constant and equal to kq/R', in other words there will be zero force due to this spherical shell. Now take the small sphere alone. A homogeneously charged smaller sphere R produces potential on its surface kq/R and everywhere outside it (r>R) the potential is kq/r. What will happen, if we put the two together, small ball inside the empty space of the larger shell (overlapping centers of symmetry of the two)? We have just said that the larger shell does not produce any force on anything inside it. This statement is independent of whether the shell is empty, or if it contains the small ball. Therefore, anywhere for r<R' the electrical potential, intensity and force will act exactly the same way, as it there was NO LARGE SHELL around! As I said, this is the Gauss law, which is at the same time one of the Maxwell's equations.

In your derivations after this point you wrongly assumed that the out-lying charge has some influence on what is happening inside it, so it cannot be correct. Nevertheless, there a little more to point out:
[Q]...So the total potential of the middle sphere is :
0 + (-kq)/c = -kq/c.[/Q]
This is perhaps a consequence of your incorrect assumption, but there is one thing you are completely ignoring here: the middle shell is grounded! This does NOT mean that its potential is 0, it means that its potential will be the same as would be the potential of the empty space at the same radius, due to the inner ball - and ONLY the inner ball. And that is simply kq/b, just as if the inner sphere was a point charge of +q. This is the fact you can build upon and when you do that, you will calculate how much extra charge has to flow into the middle shell in order to obey this prescription for the pontential. The solution of the rest of the problem (charge distribution on outer shell's surfaces) follows straightforwardly from there.

I am sorry I cannot answer any better. The Gauss law can be easily proven, if that troubles you. The role of grounding in electrostatics is just what I wrote (several time now), i.e. it's a reservoir of charge. Electrons move through the wire between the electrodes no matter what shape it has. Their motion is started by the fact that one electrode is energetically lower than the other, so there is a net force, and so the electrons just flow like water into a lower-lying barrel.

Good luck,
Daniel