I Proof of Q=CV for arbitrarily shaped capacitors

[jkfjbw]

What is a proof of the formula Q=CV for a capacitor with arbitrary but unchanging shape where C is a constant?

 

[nasu]

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.

 

[jkfjbw]

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.

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.

 

[berkeman]

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.

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]

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?

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]

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]

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.

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?

 

[jkfjbw]

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.

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.

I have taken all the electromagnetism classes needed for an undergraduate degree in physics.

 

[jkfjbw]

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.

Yes, I agree that if you do not consider the capacitance to be a constant then there is nothing to prove, however all of my physics E&M textbooks claim that it is a constant, and this is what I am trying to prove.

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?

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.

 

[vanhees71]

The surface-charge density at the plate is proportional to the jump of the electric field's normal component, which is proportional to the voltage at the capacitor (potential difference between the plates) and thus no matter how the geometry of the capacitor looks you have $Q \propto U$, and $C=Q/U$.

 

[nasu]

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.

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?

 

[alan123hk]

@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.

 

[nasu]

@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.

And this is why the superposition works. Or how it works.

 

[nasu]

But this assumes field in vacuum or a linear medium. For nonlinear medium the dielectric constant depends on the field and so, the capacitance may not be constant. So you can "prove" that the capacitance is constant for these cases when it is not variable. But these cases are useful enough to make the definition useful too. So we are somehow back where we started. But on a higher loop of the helix of progress. As a mathematician you would possibly prove that "there is at least one instance of capacity being constant".

 

[jkfjbw]

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?

 

[DaveE]

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.

 

[jbriggs444]

Suppose that we have a flat plate capacitor where the plates are separated by a dielectric. This dielectric is made of a substance that is populated by bipolar molecules. As a field manifests between the plates, the molecules tend to line up more and more uniformly. The effect is to increase the capacitance of the arrangement as compared to the same plates separated by a vacuum.

As the charge on the this capacitor increases, the electric field intensifies and the dielectric may become saturated. All of the molecules are aligned. They cannot increase their alignment any further. The marginal capacitance of the assembly decreases as a result.

The above is primarily speculation on my part, however, see:

https://physics.stackexchange.com/questions/4436/what-is-the-mechanism-of-dielectric-saturation:

"It is known from experiments that the dielectric constant of a solvent might decrease in regions where there is a strong electric field"

 

[bob012345]

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?

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.

 

[jkfjbw]

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.

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.

 

[berkeman]

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.

Sure, good point, and I use that effect in some of my analog circuit designs occasionally.

But that doesn't really relate Charge to Voltage and Capacitance in the context of what the OP is asking, I don't think.

 

[berkeman]

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.

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...

 

[Nugatory]

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.
(I would be surprised to find that there isn't a clever proof, based on the linearity of electrostatics, that capacitance is a constant - but right now you're asking for that and I can't produce it)

Another example: For centuries the assertion that energy was a conserved quantity here was no more than what you're calling "handwaving", a conjecture supported by overwhelming observational evidence but not proven from first principles. It wasn't until early last century that Noether's theorem put energy conservation on a sound mathematical basis; and that should not and did not stop us from treating it as a proven fact.

 

[DaveE]

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.

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.

BTW I did read the previous posts including the excellent link from @alan123hk in post#12:

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.

I think you won't be satisfied unless you ask a more specific question, like an EM homework problem sort of format. The answers to your question are in the math, I think. Things like linearity and superposition, potentials and conservative fields, etc.

 

[nasu]

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?

What do you mean by "charge distribution function"? What is this function for a parallel plate capacitor?

 

[bob012345]

@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?

 

[jkfjbw]

What do you mean by "charge distribution function"? What is this function for a parallel plate capacitor?

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.

 

[jkfjbw]

@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?

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]

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.

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]

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.

Yes, I was.

 

[bob012345]

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.

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.

 

[alan123hk]

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?

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.

I think this logical inference process should not be wrong.

https://www.physicsforums.com/threa...es-not-depend-on-charge.1006758/#post-6536918

 

[vanhees71]

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.

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.

 

[jkfjbw]

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.

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.

 

[jkfjbw]

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.

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.

 

[alan123hk]

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.

For capacitors of arbitrary shape, when the applied voltage is doubled, the charge density at any position on the conductor surface doubles, as shown in the figure below.

Note that $~E_1~$and$~E_2~$ are the electric field strengths at the conductor surface, their vector directions are perpendicular to the conductor surface. Also, of course, an accurate potential distribution can only be obtained by solving the Laplace′ s equation $~\nabla^2V=0~$(source-free region). You can first solve the Laplace equation, then calculate the electric field on the conductor surface based on the potential distribution, and then calculate the charge density on the conductor surface based on this electric field.

Therefore, when the voltage applied to the capacitor changes, the surface charge density at any point of the conductor and the total charge of the entire conductor change in the same proportion.

Surface Charge $~~~~~~~\rho(x,y,z)~\to~c\cdot\rho(x,y,z)$
Total Charge $~~~~~~~ \iint \rho(x,y,z) dS~\to~c\iint \rho(x,y,z) dS$

If you are still unconvinced or in doubt, maybe you need to study the properties of the solution of Laplace′s equation under certain boundary conditions, including uniqueness, linearity, etc. But I'm actually very bad at complex and esoteric math myself.

 

[hutchphd]

I think this assessment is incorrect.

Several hundred years of science would indicate otherwise.
Do you believe that Maxwell's equations are correct? Then conductors form equipotential boundaries for Poisson's eqn whose uniqueness properties are well known.

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.

I believe this work has already been done. If there is a reason it is incorrect, then the onus is upon you to point out the error. The fact that you don't see how it can be true is not really sufficient nor apparently easily correctable.

.

 

[hutchphd]

Perhaps a look at the capacitance matrix would help. I can't find the reference I wanted...maybe wikipedia.

 

[jkfjbw]

Several hundred years of science would indicate otherwise.
Do you believe that Maxwell's equations are correct? Then conductors form equipotential boundaries for Poisson's eqn whose uniqueness properties are well known.

Poisson's equation gives uniqueness of the potential provided the charge density function is given, however the situation in question is comparing the two different charge density functions resulting from depositing two different charges on the same conductors. Poisson's equation does not directly say what happens in that case. Maybe there is a way to use Poisson's equation in an argument to say what would happen in that case, but it is not directly specified by the uniqueness theorem.

 

[hutchphd]

You are not at liberty to specify the distribution of the charges on a conductor. It is maintenance of the the equipotential surface that determines the unique distribution.
Show me a counterexample.

 

[Delta2]

I don't understand the introduction of the external brought in charge at first place. When we say $C=Q/V$ depends only on the geometry of the capacitor and therefore is constant, we make the silent assumption that the voltage difference of the plates are solely due to the charges of the plates , and assume no other external charges.

 

[vanhees71]

It's just overcomplicating the in principle simple "standard model" of a dielectric in terms of linear-response theory. By definition in a dielectric you have only bound charges and thus applying an electrostatic field the positive and negative charges get a bit displaced relative to each other, leading to a polarization density $\vec{P}$ within the medium. The polarization is equivalent to a charge distribution
$$\rho_{\text{mat}}(\vec{r})=-\vec{\nabla} \cdot \vec{P}.$$
The total electric field is given by the external field and the field due to the polarization, fulfilling the equation
$$\vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho_{\text{tot}} = \frac{1}{\epsilon_0} (\rho_{\text{ext}}+\rho_{\text{mat}}) = \frac{1}{\epsilon_0} (\rho_{\text{ext}}-\vec{\nabla} \cdot \vec{P}).$$
Usually one introduces the auxiliary field $\vec{D}$ such that
$$\vec{\nabla} \cdot \vec{D} = \rho_{\text{ext}}.$$
Then you have
$$\epsilon_0 \vec{\nabla} \cdot (\vec{E}+\vec{P})=\rho_{\text{ext}} \; \Rightarrow \; \vec{D}=\epsilon_0 (\vec{E}+\vec{P}).$$
In linear response you have
$$\vec{P}(\vec{r})=\epsilon_0 \chi \vec{E}(\vec{r})$$
and thus
$$\vec{D}=(1+\chi) \epsilon_0 \vec{E} = \epsilon_0 \epsilon_{\text{rel}} \vec{E}, \quad \epsilon_{\text{rel}}=1+\chi.$$

 

[jkfjbw]

I don't understand the introduction of the external brought in charge at first place. When we say $C=Q/V$ depends only on the geometry of the capacitor and therefore is constant, we make the silent assumption that the voltage difference of the plates are solely due to the charges of the plates , and assume no other external charges.

I agree, however it was asked of me to "Forget the word capacitance for a moment", therefore divorcing the rest of bob012345's stated scenario from the context of capacitance and the expectations that accompany it.

 

[jkfjbw]

For capacitors of arbitrary shape, when the applied voltage is doubled, the charge density at any position on the conductor surface doubles, as shown in the figure below.

Note that $~E_1~$and$~E_2~$ are the electric field strengths at the conductor surface, their vector directions are perpendicular to the conductor surface. Also, of course, an accurate potential distribution can only be obtained by solving the Laplace′ s equation $~\nabla^2V=0~$(source-free region). You can first solve the Laplace equation, then calculate the electric field on the conductor surface based on the potential distribution, and then calculate the charge density on the conductor surface based on this electric field.

View attachment 301473​

Therefore, when the voltage applied to the capacitor changes, the surface charge density at any point of the conductor and the total charge of the entire conductor change in the same proportion.

Surface Charge $~~~~~~~\rho(x,y,z)~\to~c\cdot\rho(x,y,z)$
Total Charge $~~~~~~~ \iint \rho(x,y,z) dS~\to~c\iint \rho(x,y,z) dS$

If you are still unconvinced or in doubt, maybe you need to study the properties of the solution of Laplace′s equation under certain boundary conditions, including uniqueness, linearity, etc. But I'm actually very bad at complex and esoteric math myself.

How did you get $\rho=E\epsilon_0$? Is this supposed to be the differential form of Gauss' law? You'd get infinity for the $\nabla\cdot\vec{E}$ due to the surface charge.

 

[hutchphd]

This follows from Gauss's law at the surface of any continuous conductor. As one gets close enough to the surface it is locally flat and you make the usual Gaussian pillbox. A useful result.

 

[alan123hk]

How did you get ρ=Eϵ0? Is this supposed to be the differential form of Gauss' law? You'd get infinity for the ∇⋅E→ due to the surface charge.

Drawings are sometimes difficult to express in detail and can therefore be exaggerated and crude.
Of course, you need to calculate the electric field with the potential change very close to the conductor surface, and then calculate the conductor's surface charge density with the electric field very close to the conductor surface
This requires an analytical solution of the Laplace equation, or the use of a computer to obtain a very detailed and accurate numerical solution.

 

[vanhees71]

How did you get $\rho=E\epsilon_0$? Is this supposed to be the differential form of Gauss' law? You'd get infinity for the $\nabla\cdot\vec{E}$ due to the surface charge.

Along such a discontinuity the usual divergence is of course undefined, i.e., it diverges ;-)). That's clear from the invariant definition of the divergence as a limit of a surface integral over the enclosed volume. With a surface-charge density you can make the volume as small as you want without changing the enclosed charge: Just make the volume a little cube with two faces parallel to the surface that carries the charge and make the height of this cube smaller and smaller.

In such cases what is defined is the surface divergence, which is defined as the limit of the surface integral devided by the surface of the pieces parallel to the surface of discontinuity. The result is that the jump of the normal compoment of the field $\vec{D}$ across this surface is the surface charge.

This leads to the definition of the surface divergence: If we separate the space around a point $\vec{x}$ of the surface into side 1 and 2 and let's point $\vec{n}$ such that it points into region one, we define
$$\mathrm{Div} \vec{D}(\vec{x})=\vec{n} \cdot (\vec{D}_1(\vec{x})-\vec{D}_2(\vec{x}),$$
where $\vec{D}_1(\vec{x})$ is the value of $\vec{D}$ when taking the limit towards $\vec{x}$ from region 1 and correspondingly for $\vec{D}_2$. In electrodynamics with the usual meaning of the field $\vec{D}$ you then have
$$\mathrm{Div} \vec{D}=\sigma_Q,$$
where $\sigma_Q$ is the surface-charge density defined along the surface.

A related idea is the surface rotation, which you need, e.g., to express Ampere's Law in local form if there is a surface current $\vec{K}$. This leads to the definition
$$\text{Rot} \vec{H}=\vec{n} \times (\vec{H}_1-\vec{H}_2).$$
Then the local form of Ampere's law at presence of a surface current reads
$$\text{Rot} \vec{H}=\vec{K}.$$

 

[Dadface]

The capacitance of any system can be described by the equation:
C= εf
In this equation f stands for a function that depends on the geometry of the capacitor or conductor. For example with a charged sphere of radius r the capacitance is given by:
C = 4πεr
Looking at the equations this way shows that if the geometry remains constant so does the capacitance. However, any equations that you come across can break down, particularly at high voltages, due to effects such as corona discharge or dielectric breakdown.

But I think the best way to prove the constancy is to look up or do the experiments. Of course you can't test everything.

 

[alan123hk]

In fact, many laws of physics are defined by experimental results, and there may be no more underlying theory to prove them.

For example, if someone asks how to prove that Newton's second law holds for objects of any mass, size and shape? Experimenting with objects of all different masses, sizes, and shapes is impossible, so you can only trust this conclusion to be true until you find exceptions.

However, with the emergence of general relativity, I don't know if it can be proved by general relativity. It's a pity that general relativity is far beyond my knowledge and ability.

 

[hutchphd]

Earlier #37 I alluded to the "Capacitance Matrix". What I meant was ( Maxwell) the matrix of coefficients of potential.
For any collection of n conductors the relationship $$\Phi_i=P_{ij} Q_j~~where~~~~~1\leq i,j\leq n$$ and the coefficients of potential depend upon geometry.

 

[Herman Trivilino]

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.

$Q=CV$ is a definition. It is not an assertion that $C$ is constant. That is a separate assertion.