Capacitance is said to be a constant, and is defined as \$C := \frac{q}{V}\$. I have been unable, however, to find an implementation-independent way to show that charge and voltage are directly proportional to confirm this.

Are charge and voltage in a capacitor always directly proportional? If so, how can we show this without assuming the capacitor's structure (e.g. parallel plate)?

  • \$\begingroup\$ There are non ideal effects that screws with things like DC bias and dieelectric absorption \$\endgroup\$
    – DKNguyen
    Dec 12, 2019 at 2:46
  • \$\begingroup\$ A capacitor real model is more complex with ESR and leakage R and L (ESL) and complex permittivity \$\endgroup\$ Dec 12, 2019 at 2:48
  • 3
    \$\begingroup\$ You can't demonstrate a definition. That definition was chosen because charge and voltage are proportional over a wide range of values and constructions. Similarly, resistance is defined as the ratio of voltage and current because this ratio remains constant over a wide range of values and resistive-type structures. \$\endgroup\$
    – Barry
    Dec 12, 2019 at 2:48
  • \$\begingroup\$ the parallel plates can be flat, radial, monolithic and multilayer for example, so no assumption can be made. It must be defined by specs \$\endgroup\$ Dec 12, 2019 at 2:50
  • 2
    \$\begingroup\$ " We find that the voltage is proportional to the charge. Such a proportionality between V and Q is found for any two conductors in space if there is a plus charge on one and an equal minus charge on the other. The potential difference between them—that is, the voltage—will be proportional to the charge. (We are assuming that there are no other charges around.) Why this proportionality? Just the superposition principle." from The Feynman Lectures on Physics vol. 2 Chapter 6 section 10. feynmanlectures.caltech.edu/II_06.html \$\endgroup\$ Dec 12, 2019 at 8:46

3 Answers 3


You have it the wrong way round.

When charge is proportional to voltage, we say the capacitance is constant.

Capacitance is constant, or nearly so, for rigid structures made from simple materials like metal conductors and low k dielectrics. When you buy a capacitor from a catalogue, they generally have metal electrodes. There's a huge choice of dielectrics, low k ones like plastic polymers and alli or tantalum oxides, and high k ones like exotic oxides of barium, titanium, and a whole bunch of other secret goodies.

If the structure is not rigid, like the electrodes in an electrostatic loudspeaker, then the capacitance varies as the electrodes move, which would cause audio distortion if not mitigated. A tuning capacitor made of interleaved vanes is also not rigid, it's designed to be varied.

If the conductors are not simple metals, but semiconductors in a junction making a diode, then the effective edge of the conductor varies in position as the voltage on the junction varies. This is the structure of a varactor, capacitance changes with voltage.

If the dielectrics are low k, then their polarisation per electric field is generally more or less linear all the way up to electric breakdown, resulting in a constant capacitance. High k ceramics are very polarisable in low fields, but in higher fields become less so, without breaking down. This means that capacitors can be built having a high working voltage, but a reduced capacitance at that voltage, sometimes down to 10% of their zero bias capacitance. Take great care when choosing ceramic capacitors to ensure that you will have enough capacitance at the bias voltage you want to operate at, you will need to dig beyond the shortform specifications to find out.


What you're asking about is whether a capacitor is linear. That is, whether the differential capacitance \$C'=\frac{dQ}{dV}\$ is constant or whether it depends on \$V\$.

Whether the capacitor has linear behavior is more likely to depend on the properties of the dielectric material between the plates than on the shape of the plates.

If the dielectric material is linear (polarization is linearly dependent on applied electric field strength) then the capacitor will be linear up to the point of dielectric breakdown.

The proof of this is that determining the capacitance of a given structure means solving the electrostatic Poisson's equation for the potential in the dielectric, subject to the boundary conditions imposed by the shape of the plates, and Poisson's equation is a linear equation.

[Note: This is definitely true if the material is linear, isotropic (the direction of polarization is the same as the applied field), and homogeneous. I believe it is also true for anisotropic and inhomogeneous dielectrics, but I don't have the proof of it ready to hand]

If the dielectric is nonlinear, then the shape of the plates could affect how the overall capacitor structure is nonlinear, because one shape might concentrate stronger fields in a limited area where another might have more uniform field strength for a given overall capacitance.

Note: I should probably have included the condition that the plate geometry should be fixed (not dependent on applied voltage). If the plates can actually move due to the electrostatic force between them, then the capacitor will probably not be linear. Or if, like in a reverse-biased diode which we often use as a nonlinear capacitor, the charge on the "plates" is not free to move within the plates so that the effective plate separation depends on bias, you could again have a nonlinear capacitor.


Capacitance is said to be a constant, and is defined as \$C := > \frac{q}{V}\$. I have been unable, however, to find an implementation-independent way to show that charge and voltage are directly proportional to confirm this.

Are charge and voltage in a capacitor always directly proportional? If so, how can we show this without assuming the capacitor's structure (e.g. parallel plate)?

This is probably as good a time as any to actually understand dielectrics used in capacitors.

As with almost everything in nature, it's likely that when we perceive linearity in something it's really just over a small range of reality that we are exposed to and is really, from a sufficiently broad perspective, quite a non-linear behavior. For an exact perspective which makes no locally averaged assumptions, \$C_{\text{V}_\text{C}}=\frac{\text{d}\,Q_{\text{V}_\text{C}}}{\text{d}\, V_\text{C}}\$. In this form, you are admitting that the capacitance may itself vary. And that's the truer case (except where the dielectric is a vacuum, as a vacuum is fairly consistent in its behavior with respect to applied voltages.)

Suppose you have a dielectric material between two plates. The plates (of metal) are a veritable sea of thermally agitated, free electrons. The net electric field inside a plate (remember, all plates are 3D objects, not 2D) is everywhere inside of it exactly zero. (The implication here is that if you pick any two points inside of one plate and ask, "What is the voltage difference between these two points inside the plate?", then the only possible answer is \$0\:\text{V}\$.)

Outside the plates is a different story. When you impose an electric field, \$\mathscr{E}\$ (in volts/meter units), between two metal plates with any solid material between them, there may be a tendency with the molecules in the material to be affected by that electric field and to align themselves with it. If I choose to simplify the material by treating it as a series of molecularly-thin layers, then we can imagine the capacitor and its dielectric as something like this:

capacitor with dielectric dipoles in layers and two points, a and d, inside

I've included two locations, \$a\$ and \$d\$, inside the dielectric. Location \$a\$ (perhaps at a point in space within a molecule structure) will experience the dominant contributions by the dipole, which will be very large and pointing to the right. Location \$d\$, being between layers, will experience a fringing field that is the dominant contribution and this field will be weaker and to the left.

We might like to work out the average electric field between the plates due to the polarizing material in it. And at first, this would seem to be a very difficult calculation to make.

But we can at least say something about the direction, if not the exact magnitude, as there is a nice simplification that arrives from conservation law (a symmetry.) Although the complex pattern of various \$\mathscr{E}_\text{dipoles}\$ moving around through the dielectric is complex (it is), it is much less complicated when you move outside the dielectric. Here's an approximation diagram of just the dielectric itself (I've rotated the capacitor's dielectric by \$90^\circ\$):

enter image description here

Since the molecular dipoles are stationary point-charges, the round trip path integral of the electric field due to the molecular dipoles must sum to zero.

Now, here's the magic in that. Suppose the average field inside the dielectric pointed upward, then it would be the case that the round trip integral would be non-zero. But we know this cannot be the case. So, the average field inside the dielectric must point downward in the above diagram. This is a very, very general argument that cannot be disputed, regardless of the intimate complexities that may reside within the dielectric that complicates calculation. So we know that the stronger electric fields felt by points like \$a\$ in the earlier diagram must total to a larger value when summed up with the weaker electric fields felt by points like \$d\$.

The net electric field has just been demonstrated as being induced in just such a way that it must oppose the electric field set up by the voltage between the two plates (and we know that each plate has no voltage changes throughout the plate, so they are everywhere equal within a plate.) Therefore, the net field inside weakens the electric field that would otherwise be set up between the plates. That's what a dielectric does.

The dielectric constant, in fact, is just \$K=\frac{\mathscr{E}_\text{APPLIED}}{\mathscr{E}_\text{INSULATOR}}\$. And take note that \$K\ge 1\$ because the net field between the plates must be smaller than the field otherwise set up by the plates, themselves.

The reason the capacitance increases with a weakened electric field is pretty simple. It takes less work to increase charge between the plates because the insulating dielectric weakens the electric field against which this work takes place.

But insulating dielectrics also may respond in non-linear ways to larger and smaller imposed electric fields (by the plates.) And so it isn't always the case that the capacitance is a constant. It depends upon the specific details of each insulating material.

An ideal capacitor does maintain the electric field ratio of the dielectric with respect to a vacuum as a constant. (That's its definition.) So the ideal capacitance of an ideal capacitor doesn't change with different applied voltages. Proving linearity for any specific insulating material would involve those complexities I identified above with points \$a\$ and \$d\$. You'd need to solve the 3D integrals over gradient equations which would be very difficult to work out. And then, you probably would find non-linearity if you were using any real material. But if you picked an ideal one (another ideality you'd need to define carefully), then you could use calculus to prove it.


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