
#1
May2609, 11:52 AM

P: 100

Reading Feynman, there are various examples of finding voltages, forces, et cetera, when you know the charge[s] involved. But going other way? Almost nothing. To be specific, how do you answer the following question [[NOT a homework question; self invented question]]:
A vacuum tube plate is connected to a 500 volt plate bias power supply. The plate is 1 sq cm [[1 cm high ; 1 cm circumference]], 5 mils thick. When there are no other influences on the plate, how many electrons , in Coulombs, are missing? IE, what is the charge? Harry Wertmuller 



#2
May2609, 04:31 PM

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P: 5,198

I don't know what all of the various apparti (if that is the correct word) you described are. But if you have a fixed voltage across a pair of parallel plates, the relationship between the charge on the plates, and the voltage across them is given by the capacitance of the plates. The capacitance depends only upon the geometry. For a parallelplate capacitor, I think that's quite easy to calculate. I'm pretty sure it's just a simple function of the plate gap, plate area, and the dielectric constant of whatever is in the gap. Look up the physical quantity "capacitance" for more information.




#3
May2609, 10:49 PM

P: 100

I am not talking capacitance! Feynman and all other texts have abundant coverage of capacitance. I am only talking about the relation between voltage and charge. Almost all treatments explain [[derive]] voltage [[potential]] from charge[s]. I want a charge when given a voltage. No parrallel plates!. No gap! Just one single metal object connected to a voltage. I use the example of a plate in a vacuum tube because it is a real world example. A better description:
A round tubular anode [[positive charge plate ]] in a vacuum tube is 2 cm high; 2 cm in circumference; and one tenth millimeter thick. There is no other electrically charged object in the tube for this problem. Only a 500+ voltage connected to the anode. What is the charge in coulombs on the plate? 



#4
May2709, 12:05 AM

P: 100

How to find charge given voltage?
Come to think of it, an old fashioned vacuum tube anode has considerable capacitance  opposite sides of the cylinder et cetera. So try this: How can one compute the charge on a bare wire in a vacuum; wire one millimeter in diameter; one centimeter long; connected to 1000 volts [[DC]] voltage source?




#5
May2709, 01:12 AM

P: 12

There are a few relations between charge and voltage. Usually, we just say that Charge=Capacitance*voltage, but that doesn't seem to be what you're looking for, since you're saying ignore capacitance.
That makes things harder, since it requires you to consider the field of all of the charges you're talking about. The voltage between two points is equal to the integral of E*ds over any path between the two points, and you can calculate the E at every point from the charges in your arrangement if you try hard enough using coulomb's law, or maybe using gauss's law if there's enough symmetry in the problem. 



#6
May2709, 07:33 AM

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P: 5,468

So, while an isolated bare wire has lots of charged particles, the two ends of the wire are not at different voltages. Connecting the wire across a battery (for example) to induce a voltage drop from one end to the other will cause free charges to move (current). As another example, the voltage drop across a cell membrane is about 60 mV. This voltage is maintained to allow the *controlled* flow of ions into and out of the cell. 



#7
May2709, 08:30 AM

P: 10

Ahh, he asks a very real question... It is easily answered but still has a real application... It's a quantum fusion decision, me thinks,.. but as far as this world is concerned, I'm thinking a superconductor is the closest you are likely to get to an answer.. Perpetual motion perhaps? Who knows but, just keep plugging away man, one day we all, might get lucky.




#8
May2709, 08:34 AM

P: 10

biology and chemistyry is clearly classical and as for maths? yeah who cares. 



#9
May2709, 09:56 AM

P: 100

Bobbadillio, you got the idea. Andy, you don't. I am not talking about a wire across battery terminals, just a wire floating in space with some [[unknown]] amount of charge. With a supersensitive voltmeter, we measure a voltage drop between the wire and a ground [[the space shuttle?]] to be 1000 volts.
By Bobbadillio's equation V=capacitance X Q, the question then becomes how do you determine Capacitance for our object? By using the dielectric constant of free space? So we integrate over the surface of the object using the dielectric constant? Which implies the shape of the object has considerable effect on the total charge?? [[This was one one of my sub questions  does the shape matter  counterintuitive to me.]] 



#10
May2709, 10:46 AM

P: 12

The shape of the object definitely has an impact on the charge. Consider parallel plate vs. cylindrical capacitors. You might expect that the expressions for capacitance would be similar for the two, since if the difference between A and B is very small, the dielectric will be very close to a flat plane. The expression for the capacitance of the cylinder though, involves a log term.
There's a link to a ppt slide in one of this forum's discussions of cylindrical capacitors from a while back. It involves some calculus. http://www.physicsforums.com/archive.../t146621.html 



#11
May2709, 01:10 PM

P: 100

Hi, Bobbadillio. Regrading cylindrical capacitors, thanks but no thanks. I am not interested in capacitors, only charge and voltage. Perhaps best to consider this experiment:
Two objects in space are both shown [[with ultra sensitive voltmeter]] to have the same voltage drop relative to some large "ground" [[like an asteroid or space station]]. One object is a wire, the other a small solid ball of equal mass. Do they have equal charge? How exactly do yoy apply your formula Q=capacitance X V? [[Only capacitance involved presumably is dielectric constant of pure vacuum. Got your formula backward above. Must have been catawamped.]] 



#12
May2709, 02:51 PM

P: 12

Hmm, well, you also need to know if they're conductors or not. If they're not conductors, all bets are off when calculating charge.
Regardless of conductivity, we need to know more about the geometry of the situation. Outside of the context of capacitors, the only relation between charge and voltage that I know of is that charges create fields around themselves, and the voltage difference between two points is equal to the path integral of the electric field over any path between the points. Then, since the electric field is what's being considered, the geometry of the charge distribution becomes vitally important. One nice thing about conductors is that all points on the conductor are at the same potential. This isn't necessarily true for insulators. 



#13
May2709, 03:11 PM

P: 100

Yes, assume the objects are conductors.
Your response illustrates why this "experiment" befuddles me. I would think it is undeniable that a higher voltage relative to some ground would "prove" a higher charge relative to same ground. IE, voltage implies charge. But exactly how much charge? By the brute force definition, we know that path integrals from any point on my sphere or my wire, to the "ground" station, is some given value. But we do NOT know the field that gave us the result for the path integral. And surely the mass [[and MAYBE geometry?]] of the bodies comes in somewhere....... 



#14
May2709, 03:39 PM

P: 4,664

The capacitance varies as the log of the ratio of diameters of the plate and cathode (or grid). The capacitance is about
C = 2 pi e_{0} z/Ln (b/a) where z is length, b=plate radius, and a = cathode radius (b>a). Then Q = C V = 2 pi e_{0} z V/Ln(b/a) z = 0.01 m, V = 500, b/a = 4 (guess) So Q = 200 picoCoulombs 



#15
May2709, 04:02 PM

P: 370

Why does it make sense that geometry matters? When you move charge in from infinity, you have to work through an electric field, but the electric field is created by the charge distribution. If all charges are collected on a small sphere (which acts like a point charge) the electric field is easy to calculate. But, if you have some arbitrary distributed surface, the electric field is very different and probably very hard to calculate. Remember that the like charges on the conductor will try to distribute themselves over the entire surface since they repel. Actually they will exist on the very last outer layer of the conductor. You can look up the capacitance of a single charged sphere. It is easiliy found by taking the limit of a spherical capacitor as the outer conductor goes to infinity. http://hyperphysics.phyastr.gsu.edu...capsph.html#c2 This is a case where you can easily calculate. I suspect that calculating the charge on a plate is much more difficult. You can't just take the limit of a parallel plate capacitor as the conductors are separated. Instead you need the capacitance of a surrounding infinite surface (at infinite distance) with the plate at the center. Very tricky! Another tricky thing is that the spherical conductor has a capacitance of zero if it has zero radius (point charge). Now you are really confused! Me too. I think here we have to consider that the electric field goes to infintiy at zero radius, which means infinite work to truely make a point charge. 



#16
May2709, 04:03 PM

HW Helper
P: 2,156

@HarryWertM: any time you have a conductor which has some free charge on it and is at a specified voltage with respect to some other conductor, the system of the two has a capacitance. So if you know the voltage difference, you just multiply by the capacitance to get (half of) the charge difference between the two.
The tricky part is figuring out what the capacitance is. It is a purely geometrical quantity  it depends on only the size and shape of the two objects involved (and on any dielectric material that may be between them), but except for simple cases like parallel plates, cylinders, and spheres, it's very difficult to calculate. I don't know of any generic formula for finding capacitance in terms of the shapes of the conductors involved. In principle you might be able to do it by imagining some certain amount of charge Q on one conductor and Q on the other, calculating the resulting electric field, and integrating it along any path between the two to get the voltage; then C = Q/V, although you'd have to figure out how the charge distributes itself over each surface. Normally these things are measured experimentally. Anyway, the point is, the relationship between voltage and charge is the capacitance. 



#17
May2709, 11:04 PM

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P: 5,198

Okay, now that I know the OP is asking something more fundamental about E&M theory, here is a question I'll throw out there. Given that the relationship between charge and voltage is described by Laplace's equation (which states that the laplacian of the electric potential is proportional to the volume charge density), if you only know the value of the electric potential at two points in space, is it even *possible* to arrive at a unique solution for the distribution of electric charge that leads to it? I do not think so, but I want somebody else's insight.




#18
May2809, 01:11 PM

P: 100

Excellent, instructive thread. Thanks to everyone.
Regarding Cepheid's last question: In the vaguest, most intuitive sense, two points in space "define" a family of concentric three dimensional ellipsoids [[egg shapes]], or rather, an infinite series of concentric, similar, three dimensional ellipsoids could be "associated" with two unique points in space. If a charge distribution is more complex [[less symmetric]] than such egg shapes, then it would seem the two points do not offer enough information to determine the distribution. 


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