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What is a proof of the formula Q=CV for a capacitor with arbitrary but unchanging shape where C is a constant?
Q=CV is more than just a definition. It is also an assertion that the ratio of the charge to the potential difference is a constant.This is the definition of capacitance. You don't prove a definition. The ratio between the charge stored and the potential difference is called capacitance.
Sure, but so what? As long as the voltage is not so high to cause ionization of the dielectric medium, why would that ratio change? Do you have some counterexample in mind?Q=CV is more than just a definition. It is also an assertion that the ratio of the charge to the potential difference is a constant.
I do not have a counterexample in mind, but the onus of proof is not on me to disprove the claim, rather the onus of proof is on those who claim it is true in the first place.Sure, but so what? As long as the voltage is not so high to cause ionization of the dielectric medium, why would that ratio change? Do you have some counterexample in mind?
Well, if it is not constant then the capacitance is not constant. Similar discussions were started several times about resistance in Ohm's law. For electrostatics you have it actually easier. There are at least some cases where the capacitance is constant. So the definition is useful. Even if you cannot "prove" it applies to any single case, it is still an useful definition. Nothing to prove here. Same as for resistance.Q=CV is more than just a definition. It is also an assertion that the ratio of the charge to the potential difference is a constant.
I do understand his point, but he is also wrong in making it. All of my eletromagnetism books define the capacitance as a constant value. That is what I am asking a proof for. Clearly, if you do not consider the capacitance to always be a constant value then there is nothing to prove, however that is not what is being claimed in my textbooks.You did understand the point by @nasu that it is a *definition* in EM, right? What EM classes have you had so far in university?
EDIT/ADD -- I don't mean that question about classes in an aggressive way. I'm just curious what geometries you've done the integrations for to calculate the capacitance. It seems like if you've done a few of those types of integration problems that you would feel more comfortable with what we are saying.
Well, if it is not constant then the capacitance is not constant. Similar discussions were started several times about resistance in Ohm's law. For electrostatics you have it actually easier. There are at least some cases where the capacitance is constant. So the definition is useful. Even if you cannot "prove" it applies to any single case, it is still an useful definition. Nothing to prove here. Same as for resistance.
But for electrostatics, you have that potential of a point charge is proportional to the charge and the potential of any charge distribution can be calculated by superposition. Do you have an example of charge configuration with the potential depending in a nonlinear way on the charge size? If you do, wouldn't this contradict the superposition priciple?
If you do see, consider a capacitor (or any conductor ) charged with a charge Q. The potential difference will have some value V. Now what if you have the same object charged with 2Q? You may say that the potential does not have to be 2V. But if you understand superposition, you can consider the charge 2Q as the superposition of Q and Q. We know that a charge Q on the capacitor produces a potential difference V so the effect of the two charges Q+Q will be V+V =2V. If this is not true then superposition is not true, is it?I'm not following your argument. Yes, I do see how you can calculate the potential of an arbitrary electrostatic charge distribution by superposition, but I don't follow the rest of your argument.
And this is why the superposition works. Or how it works.@jkfjbw
Maybe you can try to refer to post #9 in the link below, although this is not a strict mathematical proof, but it is quite convincing to me.
https://www.physicsforums.com/threa...es-not-depend-on-charge.1006758/#post-6536918
This is because the fundamental form of the solution to the Laplace equation with fixed boundary conditions is unique. As the potential increases, all electric fields perpendicular to the conductor surface increase in the same proportion, i.e. the total charge also increases in the same proportion.
Semiconductor junctions have PN junction space charge capacitances which are voltage dependent.Sure, but so what? As long as the voltage is not so high to cause ionization of the dielectric medium, why would that ratio change? Do you have some counterexample in mind?
This is nothing but hand-waving. Please read the prior posts. Yes, it is a definition but it is more than just a definition. There is also a claim being made.We have repeatedly observed that the classical theory of EM (i.e. Maxwell's equations) are a really, really, really good description of the world we live in. We have not seen examples of situations where this theory does not hold. As others have said in a few different ways, this is built into the theory; it IS a definition. You can not invalidated the concept of capacitance and superposition without upsetting the whole classical EM theory.
You are asking us to prove a definition that is essential to the theory that we observe to always work correctly. Given the astronomical amount of evidence that has confirmed the validity of this theory, I think the burden now rests on detractors to provide evidence that it is incorrect. The first step down that path is to fully understand Maxwell's equations, et. al.
Sure, good point, and I use that effect in some of my analog circuit designs occasionally.Semiconductor junctions have PN junction space charge capacitances which are voltage dependent.
$$C = \frac{C_0}{(1- \frac{V}{V_{bi}})^m}$$
Where ##C_0## is the zero-bias capacitance, ##V_{bi}## is the built-in voltage and ##m## is the junction grading coefficient.
It's probably more than hand-waving, but I think I understand your skepticism. I still would like you to try to think of a counterexample, though. If you can't think of one, that may indicate something...This is nothing but hand-waving. Please read the prior posts. Yes, it is a definition but it is more than just a definition. There is also a claim being made.
Well yes, but the claim is supported by observational evidence far too strong to be plausibly challenged. This is not acceptable argument in mathematics where claims must be proven from precisely stated first principles and "conjecture" is pejorative, but it is the basis of all empirical science.Yes, it is a definition but it is more than just a definition. There is also a claim being made.
Yes, hand waving, that was kind of the point. It is just a definition, nothing more. A small piece of a larger theory. But perhaps we'll agree to disagree about that.This is nothing but hand-waving. Please read the prior posts. Yes, it is a definition but it is more than just a definition. There is also a claim being made.
Maybe you can try to refer to post #9 in the link below, although this is not a strict mathematical proof, but it is quite convincing to me.
https://www.physicsforums.com/threa...es-not-depend-on-charge.1006758/#post-6536918
This is because the fundamental form of the solution to the Laplace equation with fixed boundary conditions is unique. As the potential increases, all electric fields perpendicular to the conductor surface increase in the same proportion, i.e. the total charge also increases in the same proportion.
What do you mean by "charge distribution function"? What is this function for a parallel plate capacitor?It seems that the arguments being presented rely on the charge distribution function increasing by the same constant factor that the total charge on one of the conductors is increased, but how do we know that the charge will distribute itself in that way? If this were an insulator you could simply specify that the new charge distribution function is equal to the old distribution function times the same constant factor as that relating the old charge quantity to the new charge quantity, but how do you know that same distribution also holds when the material is a conductor and the charges are free to move?
I mean the ##\rho(x,y,z)## function in Poisson's equation that gives the charge density as a function of position. I'm not sure how to describe this function for a parallel plate capacitor as the plates have surface charges and surface charges present difficulties when taking derivatives, but I was waiting to ask about this detail until the method of approach for solving the problem became more clear. Perhaps describing the surface charge would involve the Dirac delta function somehow.What do you mean by "charge distribution function"? What is this function for a parallel plate capacitor?
I agree that the potential anywhere on or in the conductor is the same, given that this is an electrostatic situation. I do not think the potential of the conductor directly correlates to the total charge on the conductor. If you have a conductor with charge ##Q_1## it will have a potential ##V_1## relative to some other point. If you consider bringing in some positive charge and placing it near the conductor, you will have increased the amount of work required to bring it close to the conductor yet the charge on the conductor has not changed. Therefore charge on the conductor does not correlate to the voltage on the conductor, at least not in the general case.@jkfjbw , Forget the word capacitance for a moment and think about how one defines the electric potential of an arbitrarily shaped conductor. Do you agree the potential anywhere on or in the conductor is the same and is directly related to the total charge on the conductor?
In reference to this discussion, only consider the potential of the conductor itself, the fixed value in or at the surface. This is what is relevant to this discussion.I agree that the potential anywhere on or in the conductor is the same, given that this is an electrostatic situation. I do not think the potential of the conductor directly correlates to the total charge on the conductor. If you have a conductor with charge ##Q_1## it will have a potential ##V_1## relative to some other point. If you consider bringing in some positive charge and placing it near the conductor, you will have increased the amount of work required to bring it close to the conductor yet the charge on the conductor has not changed. Therefore charge on the conductor does not correlate to the voltage on the conductor, at least not in the general case.
Yes, I was.In reference to this discussion, only consider the potential of the conductor itself, the fixed value in or at the surface. This is what is relevant to this discussion.
I believe the potential of the conductor does not change but there will be potential difference between the conductor and the new charge ##q## equal to the work done to bring the charge where it is divided by ##q##. And if you add ##q## to ##Q## the potential of the conductor will change accordingly.If you consider bringing in some positive charge and placing it near the conductor, you will have increased the amount of work required to bring it close to the conductor yet the charge on the conductor has not changed. Therefore charge on the conductor does not correlate to the voltage on the conductor, at least not in the general case.
When the voltage applied to the capacitor changes, the potential of the capacitor conductor and the potential of the surrounding space both change in the same proportion. Since the electric field strength is equal to the rate of change of the electric potential with respect to space, the electric field distribution on the conductor surface and the surrounding space also changes in the same proportion. Note that in the electrostatic case, the charge accumulates on the surface of the conductor, and the distribution of the charge density is proportional to the distribution of the electric field perpendicular to the surface of the conductor. That is to say, the change of the charge density distribution function just adds a simple proportionality constant, that is, the rate of change of the voltage applied to the capacitor.It seems that the arguments being presented rely on the charge distribution function increasing by the same constant factor that the total charge on one of the conductors is increased, but how do we know that the charge will distribute itself in that way? If this were an insulator you could simply specify that the new charge distribution function is equal to the old distribution function times the same constant factor as that relating the old charge quantity to the new charge quantity, but how do you know that same distribution also holds when the material is a conductor and the charges are free to move?
It's of course indeed true that for the case of capacitors filled with matter, it holds only as long as the usual linear-response theory for a dielectric is a good approximation, as has been already mentioned quite early in this discussion.We have repeatedly observed that the classical theory of EM (i.e. Maxwell's equations) are a really, really, really good description of the world we live in. We have not seen examples of situations where this theory does not hold. As others have said in a few different ways, this is built into the theory; it IS a definition. You can not invalidated the concept of capacitance and superposition without upsetting the whole classical EM theory.
You are asking us to prove a definition that is essential to the theory that we observe to always work correctly. Given the astronomical amount of evidence that has confirmed the validity of this theory, I think the burden now rests on detractors to provide evidence that it is incorrect. The first step down that path is to fully understand Maxwell's equations, et. al.
Are you saying that for any point ##P_1## on the surface of the conductor and ##P_2## an arbitrary point outside the conductor, ##V_\text{old}(P)## the initial voltage as a function in terms of a point in space and ##V_\text{new}(P)## the latter voltage as a function in terms of a point in space, that if ##V_\text{new}(P_1) = c V_\text{old}(P_1)## for some number ##c##, then ##V_\text{new}(P_2) = c V_\text{old}(P_2)## for the same constant ##c##? It is not clear to me why this would be the case.When the voltage applied to the capacitor changes, the potential of the capacitor conductor and the potential of the surrounding space both change in the same proportion.
I think this assessment is incorrect. The field of the brought-in charge permeates through all of space, so if the brought-in charge is in the vicinity of the conductor, the electric field near the conductor will be appreciably changed. If you bring in a unit test charge from infinity in order to measure the work done on the test charge, and therefore determine the voltage at some point near the conductor, the test charge will now not only have to fight against the repulsion of the charge on the conductor, but also against the repulsion of the brought-in charge. If the conductor is small enough, you can more or less ignore effects due to redistribution of charge on the conductor due to the the brought-in charge being brought in, and approximate it by a point charge. Then you would be able to say that approximately, the work done in the case with the brought-in charge would be the old work plus whatever work would be needed to bring the test charge into the proximity of the brought-in charge alone. Since the work done on the unit test charge varies between these two cases of the brought-in charge being present or not, the voltage varies between these two cases as well.I believe the potential of the conductor does not change but there will be potential difference between the conductor and the new charge ##q## equal to the work done to bring the charge where it is divided by ##q##. And if you add ##q## to ##Q## the potential of the conductor will change accordingly.
Are you saying that for any point P1 on the surface of the conductor and P2 an arbitrary point outside the conductor, Vold(P) the initial voltage as a function in terms of a point in space and Vnew(P) the latter voltage as a function in terms of a point in space, that if Vnew(P1)=cVold(P1) for some number c, then Vnew(P2)=cVold(P2) for the same constant c? It is not clear to me why this would be the case.