Capacitor and Force

These two equations are for the forces between point masses and point charges. When there are extended bodies involved things are different. For gravitational forces between extended bodies, volume integrals are involved. For the electrostatic forces, surface integrals are involved. This makes a lot of difference.

Of course you can apply Coulomb's law; in fact, you MUST use Coulomb's law. And, using it properly means integrating over the entire charge distribution.

Likewise, in the case of gravitational attraction, you MUST start with the expression for the force between two point masses. It's true that in the gravitational case the concept of a center of gravity can be used, but how do you think the location of the center of gravity is found? In extended bodies with a high degree of symmetry, such as a sphere, it's obvious where it is. But in the general case, you must integrate over the volume of the object.

I also think you're making ambiguous use of the phrase "apply a formula". It's true that in calculating the force between objects, gravitational or electrostatic, you must start with the appropriate formula for forces between point masses or charges. But, for example, suppose you wanted the gravitational force between two frustums of a cone (somewhat asymmetric bodies). They each have a center of gravity, and once found, the force between them can be accurately represented by using the distance between the centers of gravity in the point to point formula, F = G*m1*m2/r^2. But what I would call a "formula" for the force between these two particular objects would involve the dimensions of each object; the formula must be found by carrying out the volume integrals. So, it's true, the point formula "applies" (in carrying out the appropriate integral), but THE (final?) formula for the force between two extended bodies is something more than just F = G*m1*m2/r^2 in the (possibly asymmetrical) gravitational case or F = k*q1*q2/r^2 in the electrostatic case.

Indeed. The following page might be of interest to all concerned:

Consider the worked parallel plate example in:

Imagine that instead of a couple of fairly thin plates 10cm X 10 cm, we replace them with two metal cubes, 10cm on a side, separated by 1 mm. The electrostatic attraction will hardly be changed by the substitution of cubes, but the gravitational attraction will be greatly increased. And, if we extend the cubes away from the 1mm apart faces, getting long bars, we can increase the gravitational attraction even more without appreciably changing the electrostatic attraction. The mathematics is the same insofar as the point force formulas still apply to the forces between point masses and charges. But for extended bodies you must integrate, over surfaces for electrostatic forces and over volumes for gravitational forces.

General equivalency between the two cases is limited to the point force formulas.

The integrations you must perform for the two cases are NOT the same. The mathematics is the same only so far as the two point force formulas go, and they are not the end all of what must be done to solve for force between extended bodies.

Jon, please consider this simple example; perhaps then you will see why Phantom's explanation is right on. Consider two point charges, let's say equal magnitude and opposite polarity, separated by a distance x. They are attracted with force F. If I have another identical pair, they will also be attracted by F. So if the two pairs are at a distance from each other much greater than x, the total force will be 2*f. But if I bring them together, so the like charges are at the same points, the force will be 2^2 or 4 times as much. OK? It really DOES matter how in space the charges are distributed.

As Phantom says, the SAME is true for gravitational attraction; two massive plates separated by x where x is small compared with the extent of the plates will not be attracted with the same force as two point masses separated by x.

Cheers, Tom

I never said it didn't matter. But I am approximating. Even the full blown volume integerals are not exact. Its a matter of how accurate one wants and I was just trying to estimate the result. The problem is that I was calcuating the force between all charge "pairs" which is Q^2 but this is wrong because its more like Q^2/A. So my estimation was flawed and not the fact that I can't assume they are concentrated at a point. Its just that when I do that I should have realized what I was doing which was essentially making the interacting force the same for all charged particle pairs(this is not true for plates... just take the opposite corner of the opposite plate).

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if 1 and 2 are the particles that we are trying to find the force on then its simply Coulumbs force. But with 1 and 3 its also coulumbs force but the component in orthogonal to the plate is very small... I did not take this into account in my approximation. Essentially what I did was treat the force between 1 and 3 as the same as 1 and 2... hence the grossly over approxmation. Bu only looking at the forces between directly opposite charges(like 1 and 2) then its much closer and should be a lower bound.

e.g.,

1 2 3 4 5 6 7 8

and compute the forces between 1 2, 3 4, 5 6, etc.. and add them up. In this case since they are all the same(because of the approximation) its really just as if they were point charges but we have k*(Q/A)^2/r^2 = force between

1 and 2. Now adding all the forces on one plate is same as multiplying by A so that its k*Q^2/A/r^2.

This of course assumes all forces that are at angles, like 1 4, 1 6, 1 8, etc.. are 0. This is not the case but for small d most of the orthogonal components are very small.

Hmm, so what your saying is that the force of gravity due to the earth, i.e., m*g, is wrong?

What your saying is that any time someone wanted to calculate there weight they would have to compute a volume integeral?

This is simply not true. We assume the earth as a point mass and compute its force.

e.g.,

F = G*mE*m/r^2 = m*a ==> g = G*mE/r^2

which works

mE = 5.97*10^24 kg, G = 6.67*10^(-11) r = 6.356*10^(6)

and gives g = 9.8 m/s^2

So assuming points isn't a wrong idea as you and phantom are trying to make it out to be. The problem is that it needs to be applied correctly which I didn't do. Theres nothing wrong with assuming point charges for a first approximation though... don't know why you guys are trying to make it out to be bad.

Again though, if you think computing two nested volume integrals is fun then by all means go ahead and do it. I'll be waiting for your result.

Cut and pasted right out of your post of 10/12/07 5:18 AM with an obvious single letter typo fixed:

"If you have charges then you have forces. Some think that the distribution matters but it doesn't."

They are exact if the object being analyzed is a mathematical construct. If the object is real, like the earth, then we can't do a volume integral because we don't have a mathematical description of the density at every point in the interior.

What I said was:

"It's true that in the gravitational case the concept of a center of gravity can be used, but how do you think the location of the center of gravity is found? In extended bodies with a high degree of symmetry, such as a sphere, it's obvious where it is. But in the general case, you must integrate over the volume of the object."

Of course, in the case of a real object such as the earth, we can't do a volume integral because we don't have a mathematical description of the density at every point in the extended body; we must simply measure the force. Having done so, we can treat the problem as though the entire mass were concentrated at the center of gravity, as I never denied, but plainly asserted.

Because it gives answers that are wrong by orders of magnitude? That's hardly what I'd call even a first approximation.

It's not a matter of fun. It's a matter of getting the correct result. And you can find a lot of this sort of thing worked out in Roark's Formulas:

if you don't want to do it yourself, or if you want to check your own results.

>

Its nice how you take it out of context. BTW, theres nothing in the quote that says volume integerals are wrong or anything.

It says distributions don't matter and they don't because all are done the same way... by a volume integeral. Any one is conceptually the same as any other.

Now maybe I said that when you reduce the plates to a point that it doesn't matter and I guess it shouldn't have said that... but for an approximation it shouldn't as you can assume point charges and get a resonable approximation in most cases(if the two ojbects are far enough apart then it will work just fine).

Take two plates 1mx1m and put them 100000m apart. Don't you think that for all practical purposes each plate looks like a point charge to the other? Why? Because the vector between any two "charges" is approximately if the plates were pointed...

i.e., 1/100000 is very small. In the case of two plates very close one has L/r where L is very large compared to r so the ratio is large and the vectors have a wide range of angles.

Physics is used to describe reality.... everything is a proximation and there are no mathematical exacts. Hence sometimes getting an order of magnitude appromixation is good enough.

duh. I don't know what your caught up so much about volume integrals. Obviously you have to do that in any case. everything involving volume in volume integeral. thats what the hell volume integerals are for... to do calculations over volumes...

But let me see you do even a simple volume integeral that is not a special case found in a calculus book by hand. Now do two nested volume integerals and lets see how far you get.

The fact of the matter is, you can say what you want but everything in physics is a approximation... and guess what? All those volume integerals your caught up on... they were derived by approximations(that is how you prove integeration in the first placed and that is how all integerals(lebesque, stieljies, reimann, gauge, etc..) are all defined.

When the is a high degree of symmetry then one usually uses that to simplify the problem if doing it by hand.

Well, your hung up on volume integerals likes something special. Its basic calculus. But you have an easy time saying that one should use them... I challenge you to compute the total force between two parallel plates using the the full volume integerals(or even surface integerals if you want). You cannot use the approximation that the site you gave uses where the plates are infinite, i.e., where E is constant between the plates and there is no fringe effect.

I'll be waiting for your solution, in full, to the problem. Lets see how far you really get. Do you even happen to know what an elliptical integeral is? if not then your going to have a lot of fun.

Nope, it was actually only off by a factor of 2. I made a mistake in the logic which is why it was so wrong... that is not hte fault of the approximation by my fault for using bad logic.

Guess what? even the site you used to prove me wrong, which I have no problem with that specically as it showed me that I was in error, is using an approximation.

The problem I have wiht you is that you keep bring up volume integerals like its a big deal. Its not. Its basic calculus. Almost all volume integerals are intractable and initially that is what I actually did but couldn't compete the integeral... even had maple try and it came up with nothing...

but since your a genius with volume integerals I want to see you do it. (what it sounds like is you like tossing the word around but you have no experience with them.)

Actually I have enough books... but when you can show me your work for the parallel plates

Heres the integeral you have to compute, I'll write it as a surface integeral since that is more appropriate for this case.

We will assume the surface charge density is constant to make your life much easier(of course your not even going to attempt to do them... doubt you ever actually did a surface integeral in your life)

k*(Q/A)^2*(int(int((dS1 - dS2)/|dS1 - dS2|^(3/2)))

where, of course, the integeral is the differential form of coulumbs law, i.e., dF = k*dq1*dq2*r/|r|^(3/2).

It sounds to me like you don't know much about what your talking about. True you see see my problem after you found a site that did a similar example but from your first post

"

The (CV)^2 term alone has a value of 16*10^(-14), so the force can't be anywhere near 3.6N"

is just utter nonsense and says you have no idea what your talking about(because its basic algebra to see your wrong). Shit, there are two other factors involved there yet you completely ignore them like they have no effect(and it really only takes r).

But even if I ignore that as a lapse of ignorance there still ist he problem of you tossing the term volume integerals around like it means something. Believe me, I do know that any time your doing with any type of object and you are computing something over that object you must integrate(even if its a point).

But since you throw it around willy nilly it sounds like you actually never computed one or you would know that its not an easy thing and in general can only be done numerically. Now this case we have a nested integeral and so its going to be 10 orders of magnitude more difficult except in special cases where there is a high degree of symmetry(the highest being a point).

Again though, since you seem to love doing volume integerals I'd like to be shut up by having you compute the one above(for all I know you might be able to do it but I seriously doubt it). Of course I'm sure if you do do it then you will make some approximation somewhere(such as the normal force is constant everywhere) and you might actually be able to do it. (but I want to see it in the general case that I described above).

Its one thing to be able to toss these terms around but do you actually have any practical experience with them?

Again, my first approximation, when corrected, is not bad at all. (it would be worse for extremely small r of course). So I'm glad you brought what you said to my attention so I could correct the approximation but I hope you see that its not wrong(after all its an approximation). But I hope you realize that your volume integerals are not as ubiquiteous as you think. Whats the point of using the volume integerals if you cannot compute them. (sure you can do it numerically but in this case it is actually somewhat difficult because of the time complexity in the general case(since its 6 integerals))

Usually one starts with a first order approximation and moves on when they need better approximations. I have mine with my corrected approximation. Its good enough for what I am doing. Maybe later I'll try to find a better one or use the ones that already exist for constant force. In any case I want you do solve the surface integerals I have since you seem to think they are easy.

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Gee, Phantom hands you the correct answer to your rather gross misconception on a silver platter, and that's the thanks he gets? Wow, aren't YOU special!

You seem to have a bit of a comprehension problem. That is not at all what I wrote. For exactly the same reason that charge uniformly distributed on a spherical shell can be considered a concentrated point charge at the center of the sphere with respect to its interaction with charges external to the shell, gravitational attraction from mass uniformly distributed on a spherical shell may be considered as a point mass at the center of the sphere with respect to its action on masses outside the shell. THAT I have worked through with an integral a time or two; it's actually quite straightforward. Try it; you'll like it. But that does NOT work for arbitrarily distributed mass, or arbitrarily distributed charge.

...

I do integrals for myself when I want to, whether for entertainment or learning or whatever. I do integrals for my employer as necessary; I'm paid for that. Now I should do an integral for YOU because??? "I'll be waiting for your signed contract and credit references."

By the way, you DO realize that there's no need to do a volume integral, don't you?

I already admitted that I did in fact fail to look closely at the expression you had there. I mistakenly assumed that it was the correct expression for the force between parallel plates, and I knew from previous experience that the force couldn't be as large as you calculated. I knew that the (CV)^2 term would dominate (in the correct expression) in spite of the fact that there are other terms. I showed this in another post.

When did I ever say that they are easy? Any particular one might be, or it might not be, but I didn't say anything about that.

Anyway, the appropriate surface integral solution to the parallel plate problem is given in:

and there's no need for me to duplicate it.

I'm sorry to say I don't think I can help your further.

Of course you can't... Cause you probably don't even know what an integral is. That site uses an approximation. It assumes that the plates are infinite in size then chops them down. It does that to get rid of the fringe effects. This is why it is much easier to do it there way. You can do the same thing with the integrals but then IT IS AN APPROXIMATION. EVERYTHING IS AN APPROXIMATION.

The fact of the matter is, that all you did was bring to my attention that I made a mistake by showing me someone else's work that but you have not actually done any mathematical work yourself and have made several blunders that make me think you even have problems with simple algebra.

I figured you wouldn't attempt the integrals... even writing it as a surface integral has an impossible solution(well, AFAIK)... but you have tried to make it sound so easy to do with all your talk of volume integrals. My approximation and thought process was actually correct but I made a mistake in my formula... once that mistake was corrected it is actually a very decent approximation. All you did was bring that to my attention, which I'm thankful... but you have also tried to make yourself out to be someone that actually knows whats going on. I have as to yet see you do any work except search the internet and found an exercise that someone else did to demonstrate my initial formulation was in error and claim a bunch of crap that you most likely have no clue about.

There were two others that actually formulated the approximation from that site before you but I did not spend enough time to think about why my formulation was in error. There was also the problem of using ceramic caps which I was in error also. But fixing the error and taking into account how ceramic caps work fixes my formula for a rough approximation.

In any case it doesn't matter much any more... this thread isn't going anywhere anymore.

And when I do compute the integeral I'm trying to get you to do by hand it agree's almost exactly(well, about 1%) with the approximation made give by that site and others(the standard approximation of assuming F is constant).

But the main point is, I'd like to see you find the closed form solution to it by hand since, again, you seem to think computing integrals by hand is easy. Remember, its not the same on that page because they did not compute any integeral for the general case of two 2d parallel plate... or, if you think they did, then do it for arbitrary charge densities.. doesn't quite matter because you say they are easy.

Jon, I'd already thought just a little bit about the integral before looking at the web page that The Phantom pointed at and had already come to exactly the same approach that the web page mentions, using annular rings with an infinitesimal width and integrating, as the right simplifying approach. I didn't go further, as their description about the approach matched my own initial impression about it and the solution they gave appeared, without spending time, to be correct.

Do you have any reason to imagine that the author of the page didn't actually solve the integral correctly? Why also do you bring up the complicating idea of 'arbitrary charge densities' when, in the case of idealized parallel plates I'd imagine that the charge was distributed to minimize the required work and thus uniformly spread (except perhaps at the edges of the plates, where closer thinking may be required, but probably wouldn't much impact the final result?)

You mentioned you were off by a factor of two (I'm not accepting or rejecting that.) The web page that The Phantom mentioned said that folks often jump initially to imagine a result that is, in fact, a factor of two off the actual mark. Is that coincidence?

Jon

Yes I do. Becuase it says it there that it assumes E is always perpendicular to the plate... and the integral is probably impossible to solve in closed form. By making assumptions you can get away with a lot. Integrals, in general, are impossible to solve in closed form. If the geometry has some symmetry involved then you can usually simplify a great deal.

quoting from that site...(the paragraph nex tot the "field and surface charge" figure,

"since we assume that the only variation is in x, d2?/dx2 = 0."

This means they are neglecting fringe effects. Why? Because then it makes the problem possible to actually do. It takes 2 integrals instead of 4.

I am not saying he solved it correctly or not. I'm saying he made assumption, just like I originally made, to make the problem easier... which is exactly what I tried to do at the start knowing how complicated the integrals are(actually I tried to integrate it using maple but since I got no result I assumed it was probalby impossible). My assumption was for a very rough estimate based on a very weak assumption.

All I'm getting at is that he made an assumption and didn't carry out the full analysis and there is a very good reason for that. Phantom is trying to make it seem like its an easy thing to do the full blown volume or surface integrals and its not... its most likely impossible and one can only get a reasonable answer if they make some valid assumption. I made an assumption but I make a mistake in applying it. We are not really even talking about my original mistake any more but the fact that assumptions have to be made to do most of the work. In fact, almost all of physics is based on assumptions... even mathematics uses assumptions constantly.

I am not at all arguing with the method the site uses to prove anything. Hell, that site doesn't do what phantom claims it does. It doesn't carry about the full blown general case without simplifying assumptions. But Phantom claims one doesn't need to make assumptions(at least he claims that the volume integrals are what one should really work with and has said that I was wrong for not using them(in fact I did used them when I first approached this problem and quickly realized it was either impossible to to hard so I tried to make some assumptions for an estimate).

Possible, I don't know. I might have been wrong in that too. I've been making so many mistakes lately that I don't know what is right or wrong any more ;/

But he says its from reasoning about the electric field which I do not do(ok, I guess I do implicitly but he does the same in his following derivation). Even a factor of 2 is not bad though ;) I wasn't looking for exact answer(because no one will find it)... I was looking at a way to get some idea of the force. My mistake was telling me it was much larger than it was and I knew something was wrong because I did some experiments and it didn't follow. A factor of 2 isn't going to be that big a difference for the experiement because I would have seen some results... even a factor of 10 would have given something.