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Everyone does know that the surface of a conductor is at equipotential during equilibrium.

I was reading Feynman's lectures where I found this (bold)line:

Suppose that we have a situation in which a total charge Q is placed on an arbitrary conductor. Now we will not be able to say exactly where the charges are. They will spread out in some way on the surface. How can we know how the charges have distributed themselves on the surface? They must distribute themselves so that the potential of the surface is constant. If the surface were not an equipotential, there would be an electric field inside the conductor, and the charges would keep moving until it became zero.

This is a much good reasoning for the surface to be an equipotential one; if there were any region to be in higher potential, charges would flow towards them to neutralize and again make the surface equipotential.

To understand his explanation, I thought of a positively charged surface that is not in equipotential status; so there would be an electric field which would prompt the free electrons inside the conductor to go there & nullify the field to make the surface equipotential, right? But what about the positive charges that are now inside the conductor? Okay, they would by repulsion move towards the surface. But what is the GUARANTEE that they would form the equipotential surface? What really happens when they go on the surface that compels them to make an equipotential surface??

[After all, you can't say:" since you are studying electrostatics, there must be equipotential region on the surface no matter what happens; that's it"-this is what my school-teacher said when I asked him.]

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Think of potential as like potential energy. If the mobile charges are electrons and they are mostly at rest they will mostly move towards lower potential energy which is higher potential.

So think of it as like a hill with electrons free to roll down hill (higher potential) until they get to a surface of the conductor at which point they are not free to move. So, you can imagine a fence built on the surface of the conductor that keeps things from leaving. And you can imagine some hills inside.

If the hills inside were flat, then indeed you could have some electrons at the edge and everything could sit there in statics. And if the hill inside was actually flat, then things can stay there.

There is a mountain called Mount Saint Helens. It looks like a normal mountain except the top looks like it was just chopped off (it actually exploded in a huge eruption in 1980).

The edge of the top is all at the same height; things on the edge would fall (and move away) if they were free to move. This is exactly what is going on in a conductor.

Now imagine you poured water on the top. The water is free to move if the pile of water gets higher than the region around. Real water has some surface tension but the point is that it can even itself come out because a region of higher water is free to move.

The same thing happens in the conductor. If you had an isolated positive charge, it would create a positive potential about itself which is a negative potential energy around itself. This would attract electrons.

So if electrons moved away from a positive charge leaving a positive charge all by itself, that would not be creating equilibrium because it would attract more electrons to it.

Charges can move. If a charge moved from the surface to another point inside and the charge there moved to a other point inside and so for until eventually a charge near another part of the surface moves to the surface, that is all fine.

So do that. Draw a bunch of surfaces of constant potentiality in red. Then draw a bunch of curves in red that are ways orthogonal to the surfaces, these are the field lines of the electric field. Now there is no force pushing in the direction of the equipotential and the force is all in the direction of the field.

Just because there is a force in a direction doesn't mean that charges go that way; that depends on the initial velocity of the charge. But imagine a region large enough to have many charges in it, the average velocity of the mobile charges could be zero because they bump into the non mobile charges except in the direction of the electric field. There they could have a non-zero average because they give and get from the non mobile charges but they consistently gain momentum in the direction of the electric field.

So now imagine a current everywhere pointing in the direction of the electric field. If you are in electrostatics then the field lines don't make any loops. So they can start at one part of the surface and end on another part of the surface. That is exactly how charge can flow.

This is important. This is saying that if all the charges were placed on the surface in a way that wasn't an equipotential and all the charges were kept at rest and were kept there for some reason so they are there in place long enough for say the speed of light to traverse across the whole object then there would then be electrostatic fields throughout the object.

If you now let go and let all the mobile charges move inside the conductor then the field at that moment is electrostatic and pointed so that current will flow through the body so that charge on the surface flows to other parts of the surface. And it never stops anywhere inside the object.

This is so frustrating. I've basically described how if you add charge to a conductor that has no net charge on the inside then the charge moves around in such a way that it continues to have no net charge on the inside.

I'm saying that your concern about the positive charge never happens in the first place. How rude. But there is some truth. If conductors start with the property of having no net charge in the body of the conductor then they can continue to have that property. That is essential for electrostatics and all statics in general.

I talked about their being no loops that really just means we can keep magnetism out of it (loops of electric fields is associated with changing magnetic fields). But we really had the electric field line go from one part of the surface to the other because there was no non-zero charge density to terminate on.

So what if for some reason you just made a conductor or just started getting to statics. Then there might be charge in the body. Now if you still have the current be proportional to the electric field you get for instance $\vec J=\sigma \vec E$ then you can take the divergence of both sides and get $$\sigma\frac{\rho}{\epsilon_0}=\sigma\vec \nabla \cdot \vec E = \vec \nabla \cdot \vec J$$

Where we used the Maxwell equation $\dfrac{\rho}{\epsilon_0}=\vec \nabla \cdot \vec E$ and we can also take the divergence of $$\vec \nabla \times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ to get the continuity equation

$$\vec \nabla \cdot \vec J=-\epsilon_0\vec \nabla \cdot\frac{\partial \vec E}{\partial t}=-\frac{\partial \rho}{\partial t}.$$

This means we have $$\frac{\partial \rho}{\partial t}=-\vec \nabla \cdot \vec J=-\frac{\sigma}{\epsilon_0}\rho.$$

Now we have stepped out of electrostatics but that is because we are talking about how it got that way. But now we see the charge density decreases when it is positive and increases when it is negative and in fact it approaches zero exponentially. So a conductor is never perfect you can think of it as something with a super huge $\sigma$ and so it gets super close to no charge in the body in almost no time.

Now, there is another reason I didn't go straight to this. There is no law of physics that says $\vec J=\sigma \vec E$ and in fact even materials that behave approximately like that have the value of $\sigma$ depend on the frequency of the change if it changes or even just depending on the temperature (and current changes the temperature).

If you placed a huge amount of charge deep inside a conductor by say shooting electrons going at almost the speed of light at it. Then yes the conductor might heat and buckle and strain and generally take some time to get that charge to the outside if it was a tremendous amount of charge. No real material is perfect.

But to the degree that the charges follow the electric field then when there is a charge density the field lines will be pointing the direction as to attract a countering charge.

Which is what we started with. If you had an imbalanced positive charge, it will attract mobile electrons to it and reduce the charge imbalance. If you have an excess of electrons in a spot they will push the neighbour mobile charges away from it and thus reduce the density in that region that contained them all.

And the continuous charge density is always for a region that contains many charges it isn't a thing that jumps up at every proton and down at every electron.

So it approaches zero by attracting more mobile charges into it or pushing more mobile charges out of it depending on whether the field diverges there in a positive or negative way and it does that based on whether there is a positive or negative charge density there.

What did Feynman want to tell in the bold lines?

If the equipotential surface wasn't everywhere tangent to the surface then it would dip inside. Sort of like if you divided your room in two and told your sibling not to cross to your half (TV sitcom style) and your brother or sister blew a bubble. If the bubble was tangent to the imaginary surface it wasn't supposed to cross; but if it isn't tangent to that surface then you can follow the surface into your side.

So the equipotential would cross inside the conductor. Which means you can follow it inside to the body of the conductor and from there you can look at the normal to the equipotential and you'll see a field line going there. That field line either terminates on a charge density (where the electric field lines terminate) or goes to the surface of the conductor. So it has two ends either both in the surface, both on densities inside or from a density inside to the surface.

If current moves in the direction of electric fields then current can flow between the two ends. If they are both on the surface then one part of the surface will have less charge at the spot that field line hit it and the the other place the field line hit the surface gains it.

If one of the ends is inside then that charge density will approach the zero exponentially fast and the current will be causing it so an opposite charge will head to the other end of the field line. So two opposite charge densities inside could be weakening each other or a charge imbalance inside can be sending its excess to the surface. And both. A place where field lines terminate can have field lines from multiple directions coming out of it and they could end on different places.

It might help if you drew the fields for two oppositely charged point charges then draw an arbitrarily surface around then so it contains them both. Where those field lines go to each other they are discharging each other. Where those field lines hit the surface you drew is where current is liking up charge on the surface. Eventually all the charge is on the surface. That's a good example to look at.

But the example in the last paragraph doesn't explain what happens when you place charge on the surface. So draw the two charges again. Draw the field lines again. And draw a new surface, but this time draw a new surface but make sure it passes through one of them and contains the other one inside. Again, the current can follow the field lines so the charge that starts out on the surface travels through the body towards the opposite charge and some travels through the conductor from that charge to where the field line terminates on the surface and some actually travels along the surface as a surface current where the surface is tangent to the field. The two concentrations of charge discharge exponentially and you end up with current spread out on the surface.

OK. Now draw a surface where it intersects both charges on its surface. Here is an example where all the charge is in the surface. It moves through the conductor and sometimes along the surface but here is the key. The current doesn't have a nonzero divergence inside the conductor so even though current flows through the body the body never acquires a nonzero charge density. So the charge stays in the surface always even though current flows through the conductor.

Make sure you can see that. Current can flow even if there is no net charge density. So when Feynman says the charge moves around the surface he doesn't mean that there is can't be current inside. But the charge imbalance stays on the outside if there was no change imbalance inside. Charge can flow inside as long as equal amounts flow in and out of every region in the body of the conductor then no imbalance will form.

And as we saw earlier if there originally was a charge imbalance inside then it decays away exponentially. This is how an ohmic material ($\vec J=\sigma \vec E$) behaves. A different material might behaves slightly differently but if it is a good conductor it will behave similarly and you can think of a perfect conductor as an ohmic material with $\sigma=\infty$ so the charge on the surface just flash gets to that final configuration that an ohmic material would go to.

And that's how to think of a perfect conductor. Just imagine a material with $\vec J =\sigma \vec E$ and find out the state it approaches after an infinite amount of time. Then imagine your object gets super close to that in a very short time. So basically imagine that $\vec J =\sigma \vec E$ for a huge $\sigma$ even if your material isn't as simple as having $\vec J =\sigma \vec E.$

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  • $\begingroup$ How could I thank you; firstly for responding- +1; thanks. $\endgroup$ – user36790 Aug 22 '15 at 0:53
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After all, you can't say:" since you are studying electrostatics, there must be equipotential region on the surface no matter what happens; that's it"

Actually, it is almost that simple - otherwise there is a contradiction

(1) Assume the electrostatic case (the electric field is constant with time and there is no electric current, i.e., no electric charge in motion).

(2) Assume the conductor is not an equipotential which implies there is a non-zero electric field within the conductor.

(3) Since there is an electric field within the conductor, mobile charge will accelerate.

But (3) contradicts (1) thus we conclude that it cannot be the case that both (1) and (2) hold; one or both is false.

It really is that simple.


To understand his explanation, I thought of a positively charged surface that is not in equipotential status; so there would be an electric field which would prompt the free electrons inside the conductor to go there & nullify the field to make the surface equipotential, right? But what about the positive charges that are now inside the conductor?

Your reasoning isn't clear to me here.

If the conductor has net positive charge, e.g., mobile electrons have been removed from the conductor somehow, the interior of the conductor is still charge neutral if assumption (1) holds.

This must be the case since, if (1) holds, there is no electric field inside the conductor and, by Gauss' law, no (net) charge enclosed by any closed surface within the conductor.

Thus, the positive charge must be on the surface of the conductor, not inside. Further, it must be distributed in such a way that the surface is equipotential.

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  • $\begingroup$ Sir, what did Feynman tell? He told that if there were varying potential at the surface, which I've considered due to positive charge, then there must be a field inside the conductor & the mobile electrons would move in the direction of the field towards those positive charges on the surface to cancel the field; since the electrons moved from inside of the conductors, there must be some unbalanced positive charges. By mutual repulsion, they would move on the surface. Is the scenario clear now? $\endgroup$ – user36790 Aug 21 '15 at 16:55
  • $\begingroup$ I want to know, what does happen then when those charges come on the surface to make the surface equipotential?? $\endgroup$ – user36790 Aug 21 '15 at 16:56
  • $\begingroup$ Some says(eg; Griffiths), when the surface is not equipotential, the field lines must not be perpendicular; there must be a tangential component. Charges move in response to this tangential force on the surface till the tangential field gets cancelled. Now, can you explain why moving in direction of the tangential field on the surface eventually cancels the tangential component of the field?? $\endgroup$ – user36790 Aug 21 '15 at 17:05
  • $\begingroup$ @user36790, it seems likely to me that your thinking in terms of making the surface electrically neutral rather than making the surface equipotential; equipotential and electrically neutral are distinctly different notions. $\endgroup$ – Alfred Centauri Aug 21 '15 at 17:46
  • $\begingroup$ Okay, then what does Feynman want to say when he says charges move in response to the field to make it zero? $\endgroup$ – user36790 Aug 21 '15 at 17:50
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You're confusing "equipotential" with "zero potential", and I think also "electric field" with "electric potential".

The electric field at point A is the force that would be experienced by a unit charge at A. The electric potential at A is the potential energy of a unit charge at A, referenced to some arbitrary zero potential point B; it's equal to the work that would be required to move a unit charge from B to A.

Loosely speaking, the electric potential is the scalar integral of the electric field. This means that a point, or even a volume, can have a high electric potential, but no electric field.

Conductors can indeed be charged, and often are. However, the electrical field across the surface, as well as throughout the interior, will be zero; if it weren't, the charges/holes would redistribute themselves until it was. And, if the electrical field across the surface and volume is zero, then all points on the surface and throughout the volume will be at the same potential (since it takes no work to move a charge from place to place on the conductor).

(Of course, conductors can actually have different potentials across them, but that's a dynamic state due to resistance and current. We're talking about the static state here.)

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Although what follows is about an electric field in space, rather than the surface of a conductor, the ideas may help:

An equipotential surface is perpendicular to the lines of an electric field, and equidistant from the source of the field, so that the electric potential at any point on the surface is the same as at any other point on the surface. In other words, it would take the same amount of work to move a charge inward or outward from any point on the surface, regardless of the path taken. The line integrals from any points on one equipotential surface to any points on another equipotential surface in the same electric field will always be equal, regardless of the location of the points on the surfaces. It takes the same amount of work to move charges between equipotential surfaces, regardless of the path. This guarantees, by definition, that a region of greater or lesser potential must be on some other equipotential surface.

To move a charge from any point on an equipotential surface to any other point on the same surface requires no work, as work done in an electric field is defined as the difference in potential between initial and final positions. Let's say you move a particle from one point on an equipotential surface to another point on the same surface. Force is applied to accelerate the particle, so the particle no longer is in equipotential. It moves around until it stops accelerating, and then begins to decelerate. While decelerating, it's no longer in equipotential, and the force required to accelerate it is recovered during the deceleration. When the particle comes to rest, no work has been done. The net force applied was zero.

Take a look at ejrb's answer and Hash's answer to this question. I know it's not exactly similar to yours, but the ideas may be helpful.

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  • $\begingroup$ Hi, @Ernie, probably you couldn't have understood what I'm asking. I know what an equipotential surface is. What I'm telling is that if the surface were at a varying potential, there would be a field & the electrons would move in response to it, as said by Feynman. But what would happen to the unbalanced positive charges that are left behind the electrons? Surely, they would go to the surface. But what does happen then that makes the charges readjust so as to make the surface equipotential?? Even Feynman wrote that they conspire. $\endgroup$ – user36790 Aug 21 '15 at 13:45
  • $\begingroup$ I only want to know that what thing let those unbalanced positive charges to readjust or conspire to make the surface equipotential; of course it is not happening by God's magic, isn't it? $\endgroup$ – user36790 Aug 21 '15 at 13:47
  • $\begingroup$ @user36790: The answers to this similar question may be what you are looking for: physics.stackexchange.com/questions/22776/…. Also, I added a first paragraph that may be on point. I think the explanation may be found in Gauss' law. $\endgroup$ – Ernie Aug 21 '15 at 15:06
  • $\begingroup$ "As the electric field outside the conductor is everywhere perpendicular to the surface, the surface is an equipotential surface"-no, it's not because electric fields are perpendicular & thence the surface is equipotential one. Electric field didn't come first; when the surface becomes equipotential, then only the electric field becomes perpendicular to the surface. $\endgroup$ – user36790 Aug 21 '15 at 15:36
  • $\begingroup$ @user36790: I didn't mean cause-and-effect. I see what you are getting at. Look at Alan Rominger's answer to the similar question I linked in my previous comment. $\endgroup$ – Ernie Aug 21 '15 at 15:51

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