Proof of Q=CV for arbitrarily shaped capacitors

[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.

 

[Mister T]

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.

 

[hutchphd]

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

That is a definition for a single pair of object(s). Some times the really big sphere is the second object. Maxwell showed the simple relation of coefficients of Potential for and arbitrary geometrical assemblage of N conductors from which capacitances can calculated for the assemblage

 

[yucheng]

Let me just jot down a few extra things to consider

Differential capacitance (nonlinearities between capacitance and voltage, this already shows that C=QV only holds in some special cases)
Self- and mutual- capacitance
Capacitance matrix

 

[yucheng]

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 doing so only adds an extra mutual capacitance term between the new charge and the conductor under consideration, but this would not affect the self capacitance term and mutual capacitance terms with other conductors, only until the extra charge is placed on ("absorbed", integrated onto) the conductor.

Maybe more clearly, one can think of a point charge as equivalent to a spherical conductor with the same charge.

 

[Srini314]

This follows from Green’s third identity applied to Laplace’s equation.

Let’s say $\Sigma$ is the surface of an isolated perfect conductor which represents an equipotential surface where the potential takes the constant value $\phi_0$.

Suppose $G(x,y)$ is the Green’s function for Laplace’s equation that vanishes on $\Sigma$. This always exists by the properties of the Laplacian operator for arbitrary shapes $\Sigma$.

Then, one can show

$\phi(x) = \int \phi(u) \frac{\partial G(x,u)}{\partial n} dS_u$.

Here $u$ lies on $\Sigma$.

But $\phi(u) =\phi_0$.

Thus,

$\phi(x) = \phi_0 \int \frac{\partial G(x,u)}{\partial n} dS_u$.

This implies the normal derivative of $\phi$ on $\Sigma$, denoted $\phi_n(u)$ is proportional to $\phi_0$.

Finally,

$Q= -\int \phi_n(u) dS_u \propto \phi_0$

And the proportionality constant depends purely on the geometry of $\Sigma$.

In fact, one can give an explicit formula for the Capacitance as

$C = - \int \int \partial_n \partial_{n’} G(u,u’) dS_u dS_{u’}$

QED