Charge distributions vs. voltage on an infinite plate

[Durato]

Don't know if this is the right place to post it, but oh well

Homework Statement

A one-sided conductor plate with negligible thickness and infinite dimension is charged to a voltage of V via electrostatic induction.
Assuming charge is distributed evenly on the surface of the conductor, describe if and how the voltage relates to charge distribution.

Homework Equations

The Attempt at a Solution

I have no clear route to the answer. What i do know is that first of all the electric field directly at the surface in a statically charged object is perpendicular and equal everywhere and can be determined to be, using Gauss' law,

E = 2*pi*k*a, where k = 9*10^9 and a = surface charge density

Then, I need to somehow relate voltage to an electric field.
So I think of the formula -V = integral of E function in terms of x. But of course i don't have a function of E in terms of x.

Trying to find one, I use the electric field a distance Z from a charged disk with radius
R (I got it from the website
lightandmatter.com/html_books/0sn/ch10/ch10.html#Section10.2
, SECTION 10.2 example 12)

E(z component) = 2*pi*k*a*(1-Z/sqrt(R^2+Z^2)). OK, but when I integrate and include the constant of integration, the constant turns out to be the initial voltage on the plate. As Z (distance from disk) approaches 0, the integral I found approaches the constant of integration. So basically, it's telling me voltage on plate = voltage on plate! But I want voltage in terms of charge density!

It goes round and round... How do you relate charge surface density and voltage (or electric field and voltage) on an infinite plate? Something obvious?

Thxs!

 

[Dick]

Yes, it does go round and round. Because V can only be defined as a potential DIFFERENCE between two points. Sometimes, but certainly not always, you pick V=0 at infinity. You can't do that in this case, because as you seem to know, E is constant, so the difference between the plate and infinity is infinite. Notice the question says "IF and how".

 

[Durato]

What I'm really confused about is relating the physics voltage definition with that of circuit voltage. Bear with me as i explain! Oh, and just by the way, this question is related to a more 'global' word problem that's been bothering me, but that doesn't really matter right now...

It seems to me that the 'usage' of voltage in circuits vs. physical simulations is different, so switching between each is hard. Here's my thoughts.

For example, if you have a series circuit containing a battery, 2 resistors, and wires connecting them, you could measure the voltage drop across one of the resistors. Then, let's say you extend the length of the wires until, say, it totals 10 meters. You could still measure the same voltage drop across the resistor, assuming the wires have no resistance.In other words, voltage doesn't depend on the distance from the voltage reference (battery) but only on the resistance of the resistor. Now we'll look at moving from the 'circuit voltage intuition' to the physics intuition.

Say a flat conductive plate is charged via electrostatic induction by placing it close to a charge sphere (say, that of a van-de-graff) which has a net charge of +Q and a voltage of V with reference to ground (the van-de-graff's, or similar, negative side is attached to Earth). A wire is then attached from conductive plate to Earth and is then disconnected. The plate now has a net negative charge and a certain voltage relative to ground. This is the 'circuit intituition part.' We could measure voltage on van-de-graff sphere, no problem there, and intuitively the charged plate should also have a certain measurable voltage on it. Now, we switch 'intuitions'. The van-de-graff is then turned off and the sphere grounded, not affecting the charged plate in any way. Now we have a floating ground reference. Here's the question. Does moving the plate farther away from Earth affect the voltage of it relative to Earth (assuming the charge and charge distrubtion doesn't change). And if so, how? I know that if i charged up a plate to a high voltage and then moved it away from earth, then measured the voltage somehow using a voltmeter, intuitively the voltage wouldn't change, just as the voltage wouldn't change in the circuit description of voltage above. In addition, since voltage is defined as potential energy per charge, if the voltage on the plate does change (because of distance from ground, Earth), does this mean that it'll take more work to bring the plate closer to the Earth ? The reason I'm asking these questions is because the physics definition seems to imply that we must pick a 'distance' for 0 volt reference, whether it be at infinity or a certain other distance away. This is confusing when compared with circuit voltage.

So, in summary what I'm wondering about is circuit/physics definition of voltage relationship. In one (physics), 'distance' seems to matter, but while in the other, distance of point from voltage source (battery) doesn't matter, assuming zero wire resistance. Switching from the electrical definition to the physics definition causes problems in my mind.

 

[Dick]

When you measure the potential difference between two points you are measuring the integrated E field between the two points. This holds both in a circuit and in free space. In the circuit you can 'extend the wires' because you are assuming, as you say, that they have no resistance. That means there is NO E field inside of the wires. That's why you can extend them without changing the resistance. There IS an E field inside of the resistors. So inside of a given wire you can put the 'distance' for 0 volt reference any where you want because all points of the wire are at the same potential. Does that help?

 

[Durato]

Thanks, it does help, makes sense for circuits! But I'm still a little confused when it comes to the question i asked above. If you charge a plate using a van-de-graff, then you should be able to measure a voltage on the plate relative to ground (via the van-de graff's connection to ground, Earth). Let's say the plate is a distance D from the ground. If you then suddenly take the charged van sphere away, then the voltage should be the same since the distance from plate to Earth (the ground) is the same. But if you move plate away, then how does the voltage change, if it does? Does it decrease, increase?

Oh, and think of this. You said one has to have a 'ground' reference somewhere. The standard 0 volt reference (I think?) is the potential difference as measured between the object and a 1 M hydrogen half-cell. So, if one carries around this cell and a voltmeter and then measured the voltage between the half cells and an object, then it wouldn't matter what 'distance' the voltage reference is from the object's voltage. Rather, voltage would then be a matter of how much net charge there is and it's distribution on the object relative to that of the hydrogen half cell. Therefore, one could find the charge distribution on the surface of the surface of the conductive object.

 

[Durato]

Of course, I forgot that a half cell is only used to measure voltage relative to other cells. But the concept remains the same. Pick an object which can be 0 volts (say, something intuitive, like a neutral bunch of atoms?) and measure everything relative to that object.

 

[Dick]

You can't carry around a 0V reference if there are external electric fields. Look, the air has a substantial electric field, on the order of 100V/m. You don't notice this shocking (ha, ha) fact because there aren't many charge carriers around. Unless you are in the way of a lightning bolt (in which case the clouds provide a huge charge reservoir). So if I take a metal plate disconnected from everything and raise it 1m, it's potential just went up 100V (relative to ground), even though the charge on it didn't change a bit.

 

[Durato]

So, you base the fact that you can't carry around a 0 v reference based on external electric fields and there being electric field in air? What if you placed the neutral atom reference in a vacuum box and furthermore, surrounded it by a faraday cage to block static electric fields? Of course it won't block a strong static magnetic field.

Anyways, this all really sort of 'theoretical.' First of all, a lot of physics problems given can be assumed to be in a vacuum, since air electric field is only taken into account if specified (usually not). So, while your answer takes into account a metal plate being raised 1 m in air, it doesn't take into account what would happen to the voltage in the presence of nothing except the ground (doesn't have to be Earth, if this introduces new variables (like a magnetic field) that have to be taken into account; could just be a big hunk of metal) and the metal plate. Sorry if I'm being picky, but i really want to understand this in all situations.

 

[Dick]

You can argue all you want, and this is getting extremely wordy. But there is no more such a thing as a 0V reference in a box as there is a 0m above sea level reference in a box. For the same reason. I think the answer to your original problem is that the charge distribution IS NOT related to the voltage. Did you think of that?

 

[Dick]

Um, you made this question up yourself, didn't you? It doesn't belong in the homework forums. It belongs in the discussion forums. The question is bad and the arguments are bad. It has the potential to confuse students.

 

[Durato]

OK, so yes I made up this question. But I did it so that I could understand voltage better. And truly, I don't see why one can't define a 0 volts reference. Is not the standard hydrogen 1M cell a 0 voltage reference for half cell reactions?
I don't know how useful it would be, but it certainly is possible to assign some random thing to 0.

EDIT: Oh, and assigning 0 absolute volts (if you wanted to measure absolute voltage) to a neutral bunch of atoms isn't that random. It is sort of intuitive that a difference in net charge (maybe not) will create a voltage and that since there is no net charge on a neutral atom, this could be '0 V' .

And about the question of the plate and charged sphere, I do not think that this is a bad question. My logic probably is, but isn't that the whole point of asking questions about something you're confused about? You pointed out that in air, the voltage of the plate would rise even though the charge didn't change because of the electric field in the air. But, what if there is no electric field? Say, it's in a vacuum. Would the voltage change if the distance changed, even if the charge on the plate doesn't? A reasonable question if you ask me.

EDIT: or maybe voltage and charge distribution are totally unrelated, as you said?

 

[Dick]

You aren't getting me. I'll repeat this again. "There is no more such a thing as a 0V reference in a box as there is a 0m above sea level reference in a box. For the same reason." Think it over.

 

[Durato]

I realize this. What I'm speculating over is how useful one would be. And isn't the standard Hydrogen half cell considered 0 volts in measuring half cell voltage. This would, it seems, have a relationship to physics definition.

My base confusion is this. We charge a plate on earth, to a voltage, relative to ground. Then, poof! Everything dissapears and the plate is in a vacuum. Does voltage measured no longer have meaning? This is similar to the problem i was having above.

 

[Dick]

No, I don't think you do realize it. Voltmeters have two probes, not one. You put one on your plate. Where does the other one go?

 

[Durato]

The other probe goes to neutral ground. All that I'm saying is that you can define one, the only pure reference for that other voltmeter probe to go to. What I'm not saying is this should replace the voltage potential used nowadays. This type of grounding is what a lot of electronics do, they are earthed to the mighty 'Earth' ground. Earth is so big that usually an adding of charge doesn't change anything. You may ask, what's the point of this? Well, as i was analyzing with my charged plate example, if I charge it up, measure voltage relative to Earth, then everything is fine, right? But then, if i move it into space far away from our Earth (ground) without affecting charge distribution, how will this voltage number help me? Dog gone it, I could care less about the number! What i do care about is finding the force that the plate will exert on an electron x meters from the edge f the plate and y meters up. So, I thought, if we initially calculated the charge distribution when we charge it up to a certain voltage on Earth, we could easily find this force! So, when i want a 0 volt reference, in this case I want to extend one of my probes, of course with zero resistance, to Earth while the other one touches the plate. So, then I could somehow use this voltage to calculate charge distribution (if it is possible, just like calculating when Earth ground is near).

Now here's a brain teaser. I'm going to ask 'Does anyone disagree?' each time, so that it is clear that I made my point and that if someone wants to disagree with it, they have to refute this logic.

#1. In chemistry, the standard hydrogen half cell is defined as 0 volts. Any other voltage potential is measured relative to it. So, if i created a half cell with copper and copper nitrate solution and a salt bridge, then its standard voltage is would have to be relative to hydrogen half cell.
Does anyone disagree?
#2. A voltage source can be created by linking two half cells together. This half cell, in turn, has a voltage V relative to the standard half cell.
Does anyone disagree?
#3. Batteries are made by linking these voltage sources together. So a battery voltage is inherently related to the voltage source in #2, which in turn is related to the 0 volt hydrogen half cell.
Does anyone disagree?
#4. Circuits can be powered by batteries. Thus, any voltage measured in the circuit can be related to 0 volt hydrogen half cell by going back from point 3 to 1
Does anyone disagree?
#5. Circuits, along with batteries, can be used to charge capacitors. These capacitors can then be disconnected and the charge difference/voltage will stay on it, if there is no other influence on the capacitor.
Does anyone disagree?
#6. Then, we can magic the capacitor away into outer space, far from Earth ground reference, and then we come back to where we started (in other posts). We have a plate, that is nowhere near Earth, with the same charge as it had on Earth. Following the logic back from steps 5 to 1, it can be concluded that this voltage can be referenced to the chemistry 0 volt half cell, thus there is already somehow defined a zero voltage reference that one could use.
Does anyone disagree?

 

[Durato]

I think I found something on the internet related to my problem. Apparently, it describes the problem of measuring the voltage of a charged, isolated conductor. When I have time to read it I will, but here's the link for anyone interested.

ce-mag.com/archive/01/11/mrstatic.html

 

[Dick]

Does anyone disagree?

I do. You are saying the same thing over and over. And it's wrong.

 

[Dick]

I think I found something on the internet related to my problem. Apparently, it describes the problem of measuring the voltage of a charged, isolated conductor. When I have time to read it I will, but here's the link for anyone interested.

ce-mag.com/archive/01/11/mrstatic.html

You don't even have to read it, I'll save you the trouble. Just look at the pictures and note the presence of a ground in every circuit diagram. You are going to have to find something more crackpot than that.

 

[Durato]

Could you please refer to the specific fact that you disagree with in the points i listed? Making general statements like 'everything you say is wrong' is not good for the continuing of discussion. Thanks!

EDIT: I imagine we could still keep on going like this with no one understanding. Let me get this clear once and for all. I'm not saying that voltage is only with reference with one point. I'm saying that for the other point one could easily choose, if you wanted, a neutral GROUND REFERENCE for EVERYTHING. I'm not disagreeing that you shouldn't have a ground reference. This is not crackpot in any way.

 

[Dick]

Could you please refer to the specific fact that you disagree with in the points i listed? Thanks!

There is no problem with constructing reference voltage differences. But none of them define an absolute 0V. The define a potential difference. Just as a meter stick defines a gap of 1m. That's all.

 

[Durato]

I guess i should abolish the use of absolute voltage. All i want is a voltage reference that could be used everywhere and with meaning. When I mean absolute, all i mean is a standard reference. The hydrogen half cell i believe is one in chemistry, and i was trying to make the point that chemistry can eventually connects with physics, so perhaps there's already a reference.

 

[Durato]

Oh, and the post i linked wasn't meant to prove 'absolute voltage.' All i wanted was a website that could provide info into the puzzling (for me) way of finding floating voltage between two points. I tried this: I tookk my voltmeter, and measured the voltage relative to a positive 1.5 V battery and a 9 V battery. It should've been an integer, but the 'measured' voltage was less than one. In addition, it seems that a charged plate would be hard to measure because once one measured, the voltage would instantly drop.

EDIT: I also tried any other combination (yes, i realize the two terminals could've been at the same potential) In general, i only got millivolt readings. Maybe i did it wrong.

Anyways, back to the main question that i was asking, I really want to know how one would find the force that a charged plate would exert on an object. Except, i want to convert from voltage to charge distribution, since it *appears* easier this way. If i measure a voltage V on some plate, then how in the heck am i supposed to calculate the force that it exerts on something? I have heard of the image charge method, but I don't see how this applies if the charged particle is lying to the right of the plate, and not right over it. But i haven't learned the method totally yet, so what do i know.

 

[Dick]

Ok. You did a great experiment. The batteries are maintaining a fixed potential between their two terminals. But you connected two floating terminals. Nothing is maintaining any fixed potential between them. You might get a momentary reading due to static charges, but the instrument will create a current path (albeit high resistance) between the two terminals and quickly discharge them to the same potential.

To compute the force exerted by the plate on another charged object you need to compute the E field generated by the charge on the plate. The reference voltage on the plate has nothing to do with it. If you know the voltage difference between the plate and the object then you could compute the force from the gradient of the potential difference, but to do that you have to integrate the E field. There really is no shortcut.

 

[Durato]

Ok. You did a great experiment. The batteries are maintaining a fixed potential between their two terminals. But you connected two floating terminals. Nothing is maintaining any fixed potential between them. You might get a momentary reading due to static charges, but the instrument will create a current path (albeit high resistance) between the two terminals and quickly discharge them to the same potential.

Exactly. Which is why i was interested in measuring floating point voltages, which normal voltmeters can't do. I've searched online and have seen instruments that can, but whatever.

This is disappointing about the voltage. But, what's keep eating at me that since voltage is such a common reference, what's the use of it if you can't use it to describe how the voltage affects physics objects in its vicinity. Obviously, one could make a student use charge distribution in problems.But in real life situations, since this is hard to measure directly, it would seem that voltage would be easier, if it worked...

Just wondering, would the following problem i propose make 'sense' in terms of a zero voltage reference? You have to flat plates, side by side, all dimensions and distances given. One plate is charged to V volts relative to the other. The other plate is 0 volt reference. What's the force on an electron at a certain plate in space?
This problem is interesting to me because I have used infinity as a 0 volt reference, but never have I used a 0 volt reference that is a fixed distance from an object.

In addition, what's your thoughts about the zero voltage standard reference that chemistry defines (standard 1 M hydrogen). How does it relate to the physical definition of voltage?

 

[Dick]

Now that problem makes sense. If the plates are flat and close together then it's reasonable to assume that the potential gradient between them is linear. The E field is the gradient of the potential, so they will produce a constant E field between them. If the objects are shaped irregularly, or even around the edges of the plates this is not longer true. I really don't have any opinions about the hydrogen cell. But even your batteries make a reasonable voltage DIFFERENCE reference. The H cell is just another kind of battery. There is no difference between the chemical notion of voltage and that in physics.

 

[Durato]

Woops, i phrased the problem wrong. I wasn't talking about parallel plates. Imagine these two plates lying flat, side by side on, a table. That's the type of configuration i was talking about.

Technically, the hydrogen thing is not a battery, rather it's a half cell. i.e., a simple 'battery' is made out of two half cells , and other half cell's (consisting of a rod of an element, like copper, and a solution, like copper nitrate) 'standard reduction voltage' can be measured relative to this. But you're right, this is just another 'difference' in potential energy. I'll have to think about this confusion of mine and state it better later. But i think, it has mainly to do with the 'distinct' definition in physics, relative to the rather wide chemistry definition. For example, in chemistry you are given a charge of 'standard reduction voltages' and from that you can determine the total voltage of a two half cell battery. But the thing i wonder is, is it the amount of electron production/depletion rate between the anode and cathode that causes each element, however similar, to have a different voltage rating? The reason for the differences were never explained in class, and i can't help wondering how this relates to the solid definition given in physics.

 

[Dick]

If there is no symmetry that makes the problem easy, then problems in electrostatics are extremely hard. You have to integrate coulombs law over the charge distribution. Electricity is chemistry is not different from electricity in physics.

 

[Durato]

Ok, thanks, i actually found many examples in my book about calculating force using charge distribution. I would prefer voltage, maybe i'll find a method to use this, like charge image method, but i think that's all. Thanks for all the help!