jkfjbw
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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.nasu said: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?jkfjbw said: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.berkeman said: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.jkfjbw said: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.berkeman said: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.
nasu said: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.
nasu said: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?jkfjbw said: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.alan123hk said:@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.berkeman said: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.DaveE said: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.bob012345 said: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...jkfjbw said: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.jkfjbw said: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.jkfjbw said: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.
alan123hk said: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?jkfjbw said: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.nasu said: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.bob012345 said:@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.jkfjbw said: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.bob012345 said: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.jkfjbw said: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.