I How can we prove that the capacitance does not depend on charge?

[jaumzaum]

How could we prove mathematically that the capacitance does not depend on charge? I tried to find this proof in the internet but I was not able. Can you guys help me?

 

[anorlunda]

It can depend on charge in special circumstances. If it does, that would be nonlinear capacitance.

https://duckduckgo.com/?q=nonlinear+capacitance&ia=web

 

[Dale]

How could we prove mathematically that the capacitance does not depend on charge? I tried to find this proof in the internet but I was not able. Can you guys help me?

You cannot prove that in general. You must experimentally measure it. As @anorlunda pointed out, capacitors can be non-linear and usually are when pushed to the edge of their specifications.

 

[Baluncore]

How could we prove mathematically that the capacitance does not depend on charge?

Capacitance is defined as the ratio of charge to voltage; C = q / v.
C cannot be dependent on charge if, for a fixed geometry, C=dq/dv remains a constant as the charge is changed.

 

[Herman Trivilino]

How could we prove mathematically that the capacitance does not depend on charge? I tried to find this proof in the internet but I was not able. Can you guys help me?

You measure the charge on a capacitor and the voltage across it. Then you change the voltage a bit and repeat. Keep doing this until you have a good collection of data. Plot that data on a graph of the charge versus the voltage. If the resulting graph is a straight line the slope of that line equals the capacitance.

 

[hutchphd]

It can depend on charge in special circumstances. If it does, that would be nonlinear capacitance.

Annoyingly it can also depend upon recent history.
A cheap mostly analogue way to do precision measurements is a dual slope Analog to Digital Convertor which compares charge to discharge time for a known and unknown signal ( great with photodiodes in current mode). One needs to choose the Capacitor wisely, because they are not that reliably linear and without hysteresis..
Mostly this is due residual polarization in the dielectric.
.

 

[jaumzaum]

It can depend on charge in special circumstances. If it does, that would be nonlinear capacitance.

https://duckduckgo.com/?q=nonlinear+capacitance&ia=web

Thanks, I didn't know that. In the internet I found a good explanation for "What makes a capacitor non-linear":

"The short answer is dielectric material properties. These properties are affected by temperature, stress, external fields and history. Most capacitor makers use dielectric materials that are as stable as they can get. Quantum Electrodynamics predicts even a pure vacuum is non linear but these effects are very small. However, near voltage independent dielectric constants are a property of pure glass, other metal oxides and Magnesium Aluminum Titanate. Ferroelectric ceramics like potassium di hydrogen phosphate or Barium titanate are an example of voltage dependent dielectrics that are used in devices like varactors to build parametric amplifiers and oscillators, or circulators usually for microwave work. Quartz crystal is a piezoelectric material extensively used a frequency standard in oscillator design. The bottom line is that there are no truly linear capacitors."

But in general we consider most capacitors obey the law Q = constant times V for many Physics Experiments, that's also what we learn in school. Also, as the above paragraph says, a capacitor in pure vacuum would be almost 100% linear. That also occurs with many other dieletrics. Why is that? Why are capacitors linear (or almost linear) in most situations? I know we can observe this experimentally, but I would want a Mathematical/Physical explanation.

 

[DaveE]

Not a real proof, but consider this bit of hand waving:

For an infinitely big parallel plate vacuum capacitor (where there isn't any dielectric material) the electric field lines will all run perpendicular from one plate to the other. The voltage generated by extra charges that make that field is the integral of the e-field from one plate to the other. The e-field strength is proportional to the amount of charge. Maxwell's equations are linear so any increase in the field strength will produce a proportional increase in the voltage between the plates. We can also use the principle of superposition in solutions. This means that the solution V₁ for an amount of charge q₁ and the solution V₂ for q₂ can be added to give the solution V₁ + V₂ for an amount of charge q₁ + q₂.

So basically, you are asking why Maxwell's equations are linear. Which, I think, is easy to see from the equations.

BTW, I'm sure this doesn't require an infinitely wide capacitor, it's just easier to think about without the curved "fringing" fields at the ends of the plates.

Also, it's easier to think of large number of charges spread out on the plates than just one or two, so you don't have to worry about where they are, etc. This is what happens in the real world, with lots of excess electrons. The superposition argument doesn't work if the charges are significantly redistributed when the solutions are combined.

 

[alan123hk]

I understand your feelings and meanings, because I vaguely remember that I had similar questions before.

Now I think about this question again. Of course, I am not capable of presenting a rigorous mathematical proof, but I can convince myself by thinking of the following deduction.

For a capacitor formed by two conductors of any shapes, its capacitance is defined as
$$ ~ C=\frac Q V= \frac {\epsilon \oint \vec E \cdot d \vec s}{ \int \vec E \cdot d \vec l } ~$$
This formula shows that the capacitance can be determined by electric field distribution in the space around the two conductors. When the potential difference between the two conductors is fixed and there is no charge density in the regions of space , we can use Laplace's equation $~ \nabla^2 \phi =0~$to find the potential distribution, and then find the electric field distribution from the gradient of the potential distribution, which is $E=- \nabla \phi$.

Now simply take a look at this Laplace equation $~ \nabla^2 \phi =0~$, when $~\phi(x,y,z)~$ is one of its solution, then $~ c \phi(x,y,z) ~$ must be its another solution, where $c$ is any constant. Please note that the $(x,y,z)$ here includes the potential on and very close to the surfaces of the two conductors and the potential distribution of the surrounding space, so I think this means that when the potential difference applied between two conductors increases according to the proportional constant $c$, the potential of all spaces must also increase in proportion to $~c~$, so the final conclusion is that all the electric field will increase in proportion to c $~ E_1(x,y,z)=- c\nabla \phi=cE(x,y,z) ~$, and the capacitance will not change. $$ ~ C= \frac {\epsilon \oint \vec E_1 \cdot d \vec s}{ \int \vec E_1 \cdot d \vec l }= \frac {c\epsilon \oint \vec E \cdot d \vec s}{ c\int \vec E \cdot d \vec l }= \frac {\epsilon \oint \vec E \cdot d \vec s}{ \int \vec E \cdot d \vec l }~$$.

 

[vanhees71]

That's of course right as long as you can treat $\epsilon$ as constant. That's exactly true only for vacuum. As soon as you have a dielectricum between the plates the linearity of the polarization with the electric field is of course an approximation (in many-body theory known as "linear response theory"), valid only for not too strong electric fields.

 

[hutchphd]

In practical terms (even without dielectric) the other assumption that will not be true in practice is that no conductor is completely rigid. Any induced deformation will yield nonlinearity particularly at high charge levels.

 

[alan123hk]

In practical terms (even without dielectric) the other assumption that will not be true in practice is that no conductor is completely rigid. Any induced deformation will yield nonlinearity particularly at high charge levels.

Very enlightening.

I want to ask another similar question. The definition of inductance is $~L=\frac \Phi I ~$, but for a closed loop of any complex shape, how to prove that the inductance of air or constant permeability magnetic core inductors will not change with the magnetic flux ?

It is also worth noting that, similar to the case of capacitors, even with air-core inductors, no conductor is completely rigid. Any deformation caused by current changes also leads to non-linearity, especially at high current levels.

 

[DaveE]

constant permeability magnetic core

No such thing, to my knowledge. Search for "B-H loop". They are very non-linear. All magnetic materials will saturate eventually.

Air isn't paramagnetic, so it should be quite linear, like a vacuum.

Many magnetic circuits take advantage of this by including an air gap to control the total permeability of the structure. The high permeability materials will effectively contain the flux even as they change value, so the total effect is dominated by a small volume of low permeability but stable material (air, paper, anything non-magnetic...).

Some cores have this air gap built in (powdered iron, MPP, etc.). They are typically small balls of paramagnetic materials sintered into a structure with many tiny air gaps.

 

[alan123hk]

No such thing, to my knowledge. Search for "B-H loop". They are very non-linear. All magnetic materials will saturate eventually.

You are right, all magnetic materials can only maintain roughly constant permeability within a fixed range.

Many magnetic circuits take advantage of this by including an air gap to control the total permeability of the structure. The high permeability materials will effectively contain the flux even as they change value, so the total effect is dominated by a small volume of low permeability but stable material (air, paper, anything non-magnetic...).
Some cores have this air gap built in (powdered iron, MPP, etc.). They are typically small balls of paramagnetic materials sintered into a structure with many tiny air gaps.

Early on, I got the information that the role of the air gap is only to prevent magnetic saturation. Later, I realized that it may have a more important role, which is to reduce the degree of non-linearity of the BH curve in the normal operating range.

 

[DaveE]

You are right, all magnetic materials can only maintain roughly constant permeability within a fixed range.Early on, I got the information that the role of the air gap is only to prevent magnetic saturation. Later, I realized that it may have a more important role, which is to reduce the degree of non-linearity of the BH curve in the normal operating range.

The air gap also allows you to store more energy in the B-field. The inductance decreases in proportion with the increase in saturation current, so $E=\frac{1}{2}LI^2$ increases at (near) saturation.

 

[jaumzaum]

Thanks guys! I think I thought of an explanation. I would be glad if you tell me if it's actually valid.

Consider a conductor consisting of N sections, where n tends to infinity. The section i has a charge Qi and a potential Vi. As we are talking about a conductor all Vi's are constant and equal to V.

 

[jaumzaum]

Thanks guys! I think I thought of an explanation. I would be glad if you tell me if it's actually valid.Consider a conductor consisting of N sections, where n tends to infinity. The section i has a charge Qi and a potential Vi. As we are talking about a conductor all Vi's are constant and equal to V.

Let's calculate the potential Vi as the sum of the potential created by all the other N-1 point charges:
$$V_i=\sum_{j \neq i} kQ_j/r_{ij}$$
Let's call $B_{ij}=k/r_{ij}$
$$V_i=\sum_{j \neq i} B_{ij}Q_j$$
As this is a linear system, we can invert it and find Q in function of V:
$$Q_i=\sum C_{ij}V_j$$
As V is constant:
$$Q_i=V \sum C_{ij}$$
If we call:
$$C=\sum C_{ij}$$
Then:
$$Q=CV$$