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I’m doing some review and some of the questions ask what would effect capacitance, and I think that because of C=Q/V that a change in Q or V should affect the capacitance yet it doesn’t. Why is that?

Qmechanic
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Bruce Wayne
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  • A change in Q or V will affect the capacitance. That is precisely what the equation says. – Steve Feb 06 '18 at 12:39
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    It is found that increasing the charge also proportionately increases the potential difference across the plates such that the ratio of the charge to voltage stays constant. – Farcher Feb 06 '18 at 12:43
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    @Steve That may be what the mathematics of the equation suggests, but that’s not the physical meaning of the equation. Instead, capacitance is taken to be constant, and depending only on the configuration of the system, not in voltages or charges. – Mike Feb 06 '18 at 12:45
  • Possible duplicate of Is mass inversely proportional to squared speed of light from relativity theory? – sammy gerbil Feb 07 '18 at 02:42
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    @Steve The equation says that a change in $Q$ or $V$ without changing the other will affect the capacitance. Equations with three or more variables simply don't answer the question "What happens to the other variables if I change one?" – Chris Feb 08 '18 at 00:22
  • @Chris, agreed. The equation says nothing about how those values would be varied independently of one another - but if you do manage to vary them independently, then capacitance will change. – Steve Feb 08 '18 at 10:17
  • @Steve The only way to vary the voltage or current independently is to force the capacitance to change. Capacitance doesn't change without some mechanical, material, or geometrical change. It doesn't respond to voltage or charge in the macroscopic world. – Bill N Feb 08 '18 at 22:19
  • @BillN, I know. It would make no sense to talk of the voltage being varied independently whilst the charge was held constant, without a change in the capacitance, because that would violate the equation! I don't know what you mean by the reference to the macroscopic world. A variable capacitor can be implemented in purely electrical terms, by switching - just as a variable voltage can, by switching the windings of a transformer (which, again, can be done electrically, not just mechanically). – Steve Feb 08 '18 at 22:44

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You have to distinguish between the simple mathematical meaning of the equation and the physical meaning. In this case the equation alone says mathematically that $C$ would change if you only changed $Q$ or $V$ but not both. But physically that's not really possible. The capacitance is taken to be constant: its value is independent of $Q$ and $V$ (though it may depend on the configuration of the capacitor), which means that this equation does not determine $C$. Instead the equation describes how $Q$ and $V$ depend on each other.

For example, this physical “law” is saying that charge can only increase if you also increase voltage. You can’t just wish charge into a capacitor. If you throw charges at it without changing the voltage at the capacitor’s terminals, the charge will flow away to the rest of the circuit to keep the voltage the same. If there is some reason the charges cannot flow away, the presence of those charges will actually change the voltage proportionally. Alternatively, the only way you can change the voltage between the capacitor’s plates is to change the plates' relative charges.

Things that change capacitance itself include changing the sizes of the plates, their separation, or the material separating them — but not just $Q$ or $V$.

Mike
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  • It is worth remembering that variable capacitors can be made, and that it is possible to increase the charge (and therefore the capacitance) independently of the voltage. But you are right that maths, when used in physics, often does not fully describe the relations and constraints of variables - and there is tacit knowledge required of what physical things the variables relate to. – Steve Feb 08 '18 at 10:21
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For a given geometry - say two parallel plates - adding more charges to the plates (more + charges on one plate, more - on the other plate) will lead to an increase in the electric field since the field between the plates is $E\sim \sigma/\epsilon$ where $\sigma$ is the surface charge density.

Since the field is constant between the plate, the potential difference $\Delta V=\sigma d/\epsilon$ where $d$ is the distance between the plates so doubling the charge also doubles the potential difference, and thus $\frac{Q}{V}$ remains constant when adding charges.

This is a general feature: adding charges will also increase the voltage difference, but always in such a way that the ratio remains constant and is determined by the geometry of the system.

ZeroTheHero
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Never forget that such formulae contain other variables as well. Other variables that may change as well, when one variable is changed.

You look at $Q=CV$ at say that changing $V$ must change $C$. But what if changing $V$ changes $C$ and $Q$? Or $C$ or $Q$? There are in fact those three options:

  1. Changing $V$ changes $C$.
  2. Changing $V$ changes $Q$.
  3. Changing $V$ changes $C$ and $Q$.

Mathematically, all these could be the case. And in reality, the second one is the case for a typical capacitor. That the second one is the case in reality is due to the physics of a capacitor (the capacitance depends on geometry and in-between material, which are fixed).

Steeven
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I would suggest looking at the simple case of the derivation of capacitance due to a parallel plate capacitor (with vacuum in-between the plates). You find $C=\epsilon_o\frac{A}{d}$. The capacitance is related entirely to the geometrical terms (A and d) and a fundamental constant of nature ($\epsilon_o$). In other words, you may think of this as a statement that $C$ depends entirely on how the capacitor is built, not the voltage between the plates.

From $Q=CV$ voltage is proportional to the change on the plates, where the constant of proportionality is capacitance. (This is somewhat analogous to stating that in an Ohmic resistor, the voltage across that resistor is proportional to the current flowing through that resistor, where the constant of propotionality is $R$.)

Bob
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Such formulas ($C=Q/V$, $v=f\lambda$ etcetera) are relations between physical quantities. In this kind of cases, the formula must not be interpreted in a sense that capacitance or speed of propagation are functions of the the quantities to the right, and certainly not that the quantities to the right are the causes of that speed or that capacitance.

It needs some experience and insight to understand this things. Teachers and textbooks should probably point this out more often.

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When we try to establish a relation between $Q$ and $V$ for a given pair of conductors, we find that $Q$ and $V$ are directly proportional to each other and can be related by a constant of proportionality.

$$Q\propto V$$ $$\Downarrow$$ $$Q=kV$$

The value of this constant, $k$, does not depend on $Q$ or $V$ because if it did we wouldn't have found $Q$ proportional to $V$ in the first place.

This constant of proportionality is defined as Capacitance $(C)$. Since we define capacitance as a constant of proportionality, it does not depend on the variables.

$$Q=CV$$

By changing $C$, you can only change the ratio of $Q$ and $V$ and this ratio will always stay the same for a given set of geometry and medium.

Mitchell
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