lol, so I guess this is how you admitt your wrong?
"Look at Jim Thompson's comment about calculating force.
For a reality check, consider that when you do electrostatics experiments you're happy to lift little bitty styrofoam balls with multi-kV of static electricity -- what in heck are you going to manage to do with a measly
20V?"
Jim Thompson was wrong and his comment about calculating force was really a question.
Then you try and give me a "reality check" that is illogical. It has nothing to do with the voltage. The high voltages in static electricity is what produces the very small charges that produce very small forces but those objects do not have any significant charge carrying ability such as a capacitor. Both your statements are illogical and wrong and have a taint of ego(your trying to feed me a load of BS like you know what your talking about). Luckily I'm not the type to roll over and play dead.
I'm still waiting for the responses of the others who didn't know shit either but tried to feed me some. I like Phantom's reply:
"The (CV)^2 term alone has a value of 16*10^(-14), so the force can't be anywhere near 3.6N"
Total BS.
I supposet he only real guy who knows somewhat what he is talking about is Bill.
Its one thing to not know what your talking about and show some modesty when replying but to not know what your talking about and pretend like you do is just sad. I expected much more from you guys ;/ You sure do act like you know it all...
I would question your force formula. Mind you, its 40 years since I thought about this stuff, but my recollection is that
F = k * Q1 *Q2 / r ^ 2 applies to two spheres in free space. The work done ( or the energy stored ) is to separate the spheres to 'infinity'.
Things are quite different with parallel plates. Look up a text on the field distribution for different configurations. Also think about exactly where the charge is stored with a dielectric in place, and where the fields go as you move the plates apart, but dont expand the dielectric.
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I'll assume all your formulas and calculations are correct. So you have a 20pF cap with your penny. This means the (tiny) force due to the charge is distributed across the entire surface area between the two pennies (by 1nF, I think you meant one nanoNewton?).
A ceramic capacitor consists of hundreds (thousands?) of layers of very thin metal bonded to very thin films of ceramic, arranged so that every second layer of metal is connected to one terminal, and the remaining layers are connected to the other terminal.
The force (assuming your calculations are correct) is still distributed across the entire (tremendous) surface area of the capacitor, which now consists of thin sheets of metal bonded to thin sheets of ceramic film, plus the area of the penny. Since almost the entire surface area is across the metal film surface, through the ceramic dielectric, almost all the charge is also there, which means almost all the force is there. You have, of course, exactly the same surface area across the surfaces of your pennies, thus exactly the same charge along their surfaces, thus exactly the same force between them as you had before. Your 3.6N is there, all right; it's distributed across the metal films inside the capacitor. The pennies still have the original picocoulombs of charge at their surfaces, thus the original nanoNewtons or less of force between them.
I'm rather shocked that this easy error wasn't caught by some of the guys who are actually more knowledgeable than I about this sort of stuff.
As soon as you try and move the plates apart it is air-gapped - ceramic dielectric is brittle. The odds are that you are looking at a multilayered structure in any event - your one millimetre could well be a stack of very much thinner capacitors, stacked in series but wired in parallel.
Agreed. Getting the right idea about what's going on usually means eliminating a lot of wrong - though superficially plausible - ideas. Nobody talks much about the process of getting rid of the bad ideas, but it always takes a while.
Then that explains the reason why it doesn't work ;) Wish someone would have mentioned that sooner ;/ I thought they were just parallel plates with a dielectric in the middle ;/
Doesn't matter how the force is distributed. If I push you with a force of
100N per hand for 50N with two hands you'll still feel an overall force of
100N. Your thinking of pressure which is F/A. Its true that the force is distributed but doesn't matter because there is a net force on the plates and this is what I was calculating. (assume the plates are points then you don't even have to worry about how they are distributed and it is a good approximation)
Now your right about the distribution of the forces. The charge Q is actually very small per plate and each plate(layer) experiences a very small force and there is no net force because all the inner layers cancel out except for the two outer ones which have a very small force(Q/n where n is the number of layers).
lol. Yeah, I'm supprised too. Most of these guys claim to be geniuses but this is simple physics. Almost every one was claiming my physics was wrong(saying that the force equation was wrong) and none actually got what I think you got.
So I think you got it. Your answer makes much more sense and most likely is the correct answer. Eventually I might try this with two very large plates, say 1m^2 with 10V and see what happens.
i.e., C = e_0*A/r, Q = C*V, F = k*Q^2/r^2. with r = 1mm F ~= 70N. Even with just 1/4m^2 plates its 4N which is much more managable.
Again, the force distributed has nothing to do with it because that is internal to the plates. Think of a bed of nails. When you lay down on one the force is exactly the same. Its your weight. The pressure is what is different. For a bed of nails its F/(number of nails*surface area of nail head). For a sheet of metal its F/Area. Thats why its never a good idea to lay on just one nail(force is the same but pressure is much larger and this pressure is enough to puncture). Sure you can say that all the force is applied to that one point but you are actually then talking about pressure.
Let me give you another example just so you get it. Suppose you have a very large object to pull. You can pull with a force F. Now you have any number of ropes. You could hook up one rope and pull wiht your force or 1000 ropes. Do you think increasing the number of ropes will make it easier to move? (for practical reasons you might use a few more ropes of course). You might say that at each point where a rope is connected there is a larger force than other places. This is not true in a rigid body because the force distributes itself throughout the body(this is not entirely true because of elasticity). Any any case usually one is only concerned with the net force and not how the forces propagates and so your force is (F/n)*n = force per rope * number of ropes used. The net force on the body is simply F and thats all that really matters if you are not concerned about internal forces.
The whole point here is that its true that there are a ton of smaller forces... that is, there is one force for each charge pair given by F = k*q^2/r^2 where q is the charge of an electron. But adding up all the forces gives F = k*Q^2/r^2 (since sum(sum(k*q_i*q_j/r^2)) = k/r^2*sum(q_i)*sum(q_j)). This total force is the force you'll be seeing.
I think John Perry hit the nail on the head. I don't think anything is wrong with my analysis except I misunderstood how a ceramic capacitor worked. I'm going to try it with just two large plates and see what happens and if it doens't work then I'll go back and look at my calculations. (I suppose I should have tried that in the first places but large plates are harder to manage)
I failed to notice that you weren't using the correct formula for the force between parallel plates. I thought you knew that Coulomb's law is for the case of point charges.
If you use the correct formula:
F = 2*pi*k*(CV)^2/A
With a plate area determined for a 20 nF cap with .001 m plate spacing and ceramic dielectric constant of 200, say, you will get a plate area of about .05 m^2.
The (CV)^2 term dominates, and you'll get a force of about .18 newtons.
As a check on your calculations, you might try calculating it in a different way. For example, if you move a plate distance dx, and the capacitance changes by dC in the process, and you still have the same charge on the capacitor, you can (1) calculate the voltage change, then (2) calculate the energy change, then (3) knowing the distance and the energy change, calculate the force. Of course, the answer should come out the same as your calculation concerning the force attracting charges. Often, a good way to discover "wrong" ideas is to try to arrive at the answer by as different a path as you reasonably can.
With regard to the force between charges which are not separated by vacuum (or, essentially, by air), beware! Consider the following: if I make an air-insulated capacitor with, say, 1000pF capacitance, and put 1 nanocoulomb of charge on it, it will be charged to 1 volt. The energy stored in it is C*(v^2)/2, or 0.5 nanojoule. If I replace the air dielectric with a high-K ceramic, let's say with a relative permittivity of 1000, then the capacitance is 1 microfarad. If I put the same 1 nanocoloumb of charge on it, the voltage will be 1 millivolt: the capacitance went up 1000 times; the voltage went down
1000:1. Now what's the stored energy? That squared voltage term has a bigger effect than the linear capacitance term: the stored energy is 1/1000 as great as with the air dielectric, only half a picojoule. THAT tells me that, since the spacing is the same, the FORCE is 1/1000 as much. So to the extent the ceramic "magnifies" the capacitance, to the same extent, it reduces the force a given charge produces.
Take that back. I recall calculating solenoid force versus position in the early '70's ;-)
...Jim Thompson
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well, all your doing is taking the derivatives then plugging them in. Its a different path to another calculation that should work out too but if my formulas are wrong in the first place or my logic is flawed then chances are I would either get the right answer but still be wrong or get the wrong answer and still be wrong ;)
I see what your saying though but it is a elementary problem and I seriously doubt my calculations are wrong(I use maple to do them and I've done them enough that any chance of it being wrong is pretty low). Now I might be misapplying the formula but again, its pretty basic. If you have charges then you have forces. Some thing that the distribution matters but it doesn't. It is very similar to gravity(in fact it is equivalent except for scale). I'm sure they don't object to treating an object as a point mass at the objects center of mass? Its always perfectly fine to do this(at least with newtons laws it works out perfectly(can be proven)).
I believe John Perry solved the problem.
Yes, I know that. Its pretty simple since Q = C*V and in F = k*Q^2/r^2 = k*(CV/r)^2. It means all factors(C,V,r) effect the force with the square. This is sorta the problem and why such a large charge, or large voltage, or very small distance must be used to get a significant force. Its not easy to get them(if it was then life would be totally different).
My assumption about ceramic caps was wrong. I didn't know it was layered.
Since one has layers the charge is distributed between each layer. Suppose there are 2n interleaved layers and charge Q. Then each layer has charge Q/n. Every layer experiences a force between every other but almost all cancel out.
By assuming all the charge Q was on just one plate seperated by a distance r was wrong(Which is how jus 2 plates would actually work).
Some initial investigation shows that the layers have a significant effect in reducing the force. No only is the charge on each plate reduced by Q/n but the distance between the first and last is increased because of all the layers inbetween. So treating a ceramic cap as a parallel plate cap is wrong as I did it. (using CV to get the charge and assuming it was on each plate because it should be Q/n but I have no idea what n is).
So there are forces there but they are much smaller than what I was thinking. For a very large parallel plate cap one has C = e*A/d, Q = C*V, F = k*(Q/d)^2 = k*(e*A/d*V/d)^2 = k*(e*A*V/d^2)^2.
For A = 1/4 m^2, V = 10V, d = 1mm, one gets F = 4N. This is without a dielectric in a vacuum. I imagine that that if one inserts a dielectric(which they will probably have to if they want to insulate it) that it will cut the charge in half because the dielectric sorta acts as a layer itself. In any case it should be an order of magnitude approximation.
Note that increasing d = 2mm gives F = 0.2N so the F is heavily dependent on d. (obviously it is d^(-4)). Changing d by a factor of 10 changes F by a factor of 10000.
OK, though I believe all that's accurate, it doesn't really match what you wrote. I believe you wrote about charging the ceramic cap to the same voltage as an air-insulated cap. So if you're scratching your head wondering what I was going off on -- don't worry, I was just off on a tangent. The paragraph before that in my earlier posting still might be helpful, though.
You seem particularly resistant to the notion that you are using the wrong formula to compute the force between the plates of a parallel plate capacitor. The formula for the force between point charges won't give the correct result. You have to divide the plates up into little dxdy pieces, with each having the appropriate little bit of the total charge on a plate, and integrate the force between each little bit of charge on one plate and all the little bits on the other plate. See:
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and:
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The second reference goes through the integration necessary, and even gives an example of plates of dimensions 10 cm X 10 cm and 1 mm spacing charged to 300 volts. Their calculated force is .00398 newtons (398 dynes).
Using your formula k*(e*A*V/d^2)^2 and plugging in their numbers:
k = 8.988E9
e = 8.854E-12
A = 10cm * 10cm = .01 m^2
V = 300
d = 1mm = .001 m
we get 6.34 newtons, considerably more than the .00398 newtons they got.
Now there, finally, is a reasonable resolution to the problem, and an explanation of just why Jon's result was incorrect. As a check, I did a d(energy)/dx calculation for the 10cm x 10cm, 1mm spacing plates charged to 300V. My capacitor calculator tells me it will be about
88.54pF, and if the plates are spaced by x, then d(energy)/dx = C*(V^2)/(2*x) -- which in this case evaluates to 3.984e-6 joules/mm, or 3.984e-3 joules/meter.
It isn't equivalent. For one thing, there is no such thing as shielding with respect to gravitation. The presence of a sheet of lead between the earth and moon would have no effect on the forces the earth exerts on the moon; the forces exerted by the sheet of lead would be additive to the earth's forces. But, if the moon and earth were oppositely charged, a conducting shield between them would alter the electrostatic forces the earth would exert on the moon.
Consider further that there is no such thing as a Faraday shield in gravitation. In electrostatics, there is no electric field inside a hollow conducting container. Even with a tiny hole in it to give access to the inside, this is essentially true, but such an effect does not exist for gravity. It's true that inside a hollow sphere made of some isotropic material, there is no gravitational field, but this is only true for a sphere. Inside a hollow non-spherical isotropic object, there is a gravitational field. But in the world of electrostatics, it doesn't matter what shape a hollow container has, there's no electric field inside. The distribution of charge on the outside matters.
Yes I am resistant because the formula applies to this situation. I might be making some mistake in applying it but that is a totally different story from using the wrong formula.
I know you can use calculus for the general case and that can be done but it shouldn't result in any significant different in this case(because of the large degree of symmetry) and actually my result *should* be a lower bound since I don't take into account any forces acting at an angle.
I do not know why the results are so significantly different. Obviously the formula I wrote above doesn't come out with any reasonable answer and is several orders of magnitude off. (it should actually be lower) I either there is a factor wrong or something else is wrong. The logic itself should work just fine and its exactly the logic that site uses except I do not worry about charges that are not directly across from any other.
I think that is my problem though. I'm computing Q^2 which gives the interaction force between every pair of charges but doens't take into account the angle at which the forces act so its treating it as a if the force is always constant for every particle pair. So actually I'm guessing what I should have is k*(Q/A)*(Q/A)/d^2*A (as an approximation.. dimensions are wrong of course)
in this case I get about 0.006N for the example above. So the real problem, like a few have mentioned is that its wrong to assume that all the charge is concentrated at a point because when I do this I'm basically saying the force is the same over a all charge pairs when its definitely not. By realizing that the force is really only that strong for one pair and then gets drastically weaker(which for my approximation I just say its 0) I get a much more reasonble approximation. But in this case it should be lower and its still higher ;/
I'll work on it some more and see if I can still apply the coulumbs law or not. I do think it can be used as a first approximation but it needs to be used properly instead of blindly like I did. (Actually my result works well if the distance between the plates is much larger than the size instead of vice versa).
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