2
$\begingroup$

My textbook states that:

...The capacitance $C$ depends only on the geometrical configuration (shape, size, separation)of the system of two conductors. [As we shall see, later, it also depends on the nature of the insulator (dielectric) separating the two conductors.]...

Now I wanted to know if these were the only two factors on which the capacitance of a conductor depends. So I tried thinking about it from the basic definition, which is:

Capacitance is the ratio of the change in electric charge of a system to the corresponding change in its electric potential. (Source: Wikipedia)

i.e., $$C= \frac {dQ}{dV}$$

where, $V$ is the potential of the conductor with respect to zero potential at infinity.

Now, I can not think of why $C = f(V)$ isn't a possible scenario, assuming that shape size, etc remains constant.

So:

  • Why cannot $C$ be a function of $V$?
Improve this question
$\endgroup$

4 Answers 4

Reset to default
4
$\begingroup$

Capacitance is a constant by definition!

Indeed, as noted in some of the answers, there exist situations where one actually prefers to define a voltage-dependent capacitance, $$ C(V) = \frac{Q}{V} \quad \text{ or } \quad C(V) = \frac{dQ}{dV} $$ just like one defines sometimes non-linear resistance or conductance. This is a popular approach in engineering.

However, in more theoretical setting one usually defines the capacitance as a the first coefficient in the Taylor expansion of charge (or sometimes in the expansion of energy near its minimum) in powers of the potential: $$ Q(V) = Q(0) + V\frac{dQ}{dV}\Biggr|_{V=0} + \frac{V^2}{2}\frac{d^2Q}{dV^2}\Biggr|_{V=0} + ... =Q_0 + CV + \dotsb $$ Thus, the correct definition of capacitance is $$ C = \frac{dQ}{dV}\Biggr|_{V=0}, $$ and it is voltage independent by definition.

The same applies when one defines resistance/conductance as a linear response coefficient in a current-voltage relation or when one defines the effective mass (as the band curvature near its minimum/maximum). But, as I have already said, occasionally one would use these terms for the parameter-dependent derivatives.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you, sir! Understood. I have an extra question: In the Taylor series of $Q(V)$ can $Q(0)$ be a non-zero entity (assuming that potential is zero at infinity). $\endgroup$
    – user249968
    Commented Oct 9, 2020 at 11:07
  • $\begingroup$ In principle, yes - by one often considers systems which are, overall, electrically neutral. For example, an electric circuit is electrically neutral, but there may be dis-balance of charge on the capacitor plates (one has $+Q$, the other $-Q$). $\endgroup$
    – Roger V.
    Commented Oct 9, 2020 at 12:21
  • $\begingroup$ The capacitance of an ideal capacitor is constant by definition. The capacitance of a practical capacitor is what it is. We talk about ideal capacitors (and other ideal components) because they obey simple mathematical laws which usually are "close enough" to the actual behavior of practical components that we can ignore the difference. But as the "Varicap" answers here point out, not all practical components are so nicely behaved. $\endgroup$
    – Solomon Slow
    Commented Oct 9, 2020 at 13:18
  • 1
    $\begingroup$ @SolomonSlow I am talking here about a capacitance as it is understood in theoretical physics - e.g., Landau&Livshits' Electrodynamics of continuous media defines capacitance of a system of conductors as an expansion coefficient of their energy. This is how many quantities are rigorously defined. I agree, that this is rather different from the definitions in basic physics or engineering. Just like Lagrangian and Hamiltonian mechanics looks very different from Newton' laws that we study in school. $\endgroup$
    – Roger V.
    Commented Oct 9, 2020 at 13:32
2
$\begingroup$

It all depends on the capacitor you use. Varicaps or "voltage controlled capacitors", are common circuit components that change their capacitance when the applied voltage changes. They are used as tuning components in oscillators and similar circuits.

Varicaps are reverse-biased diodes in which the depletion layer thickness varies with the applied voltage. The depletion layer is the dielectric of the capacitor and determines the capacity. This effect happens in all diodes, but varicaps are engineered to maximise it.

Improve this answer
$\endgroup$
2
$\begingroup$

$C$ can be a function of $V$, for example the capacitance of a varicap diode is controlled by the voltage across it.

However the voltage does not directly set the diode capacitance. What it does is to control the separation of the charge layers. It is this varying separation which results in the capacitance change.

Thus, the capacitance is a function of voltage only because the separation is a function of voltage and the capacitance is a function of that separation.

This kind of secondary effect is the only way that voltage can affect capacitance; it has to alter one of those basic physical parameters.

Improve this answer
$\endgroup$
1
$\begingroup$

Capacitance does depend on the applied voltage.

I think what the author means is that for many substances, capacitance will not change regardless of the potential differences between the plates. He could also mean that for the same voltage different substances will cause different capacitance.

Consider two plates with a certain voltage in between (we won’t change anything about the plates - shape, size and separation). Now consider keeping this voltage constant whilst we insert then remove different dielectric substances. Every time we insert one we measure the capacitance $Q/V$. We will keep getting different values of $C$ for different substances. This maybe what he means by capacitance not being a function of voltage.

But changing the voltage will change capacitance (but once again this is not true for many substances due to internal properties of these substances) and therefore $C = f(V)$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.