0
$\begingroup$

I know that the Capacitance of a capacitor does not depend on the Voltage (because it is a measure for how much charge a capacitor can hold) and it largely depends on the Area

i.e. it holds that : $C= \frac{Q}{U}$ but it also holds that : $C= \epsilon \frac{A}{d}$ where Q is charge, U is voltage, A is area, d is the distance , epsilon is the electric field constant

Is it true that the capacitance depends only on: area, and distance from the plate of eachother ?

$\endgroup$
1
$\begingroup$

For a parallel plate capacitor yes, the capacitance only depends on the plate area and plate separation.

This is a good example of knowing what your equations mean, and you are on the right track. The definition of capacitance we use $C=Q/U$ is true for a given capacitor with capacitance $C$, but it just gives a proportionality between the charge on the plates and the potential difference between them. If we wanted to rearrange the equation to show more of the "cause and effect" we would say $Q=CU$, meaning if we increase the potential difference between the plates, we know what charges will end up on the plates. However, I wouldn't say the capacitance "depends" on these things, since changing either $Q$ or $U$ does not actually change the capacitance, since if we change one the other changes in proportion according to $C$.

On the other hand, the equation $C=\epsilon A/d$ does show us actual dependencies. We can change the area and actually change the capacitance (How $Q$ and $U$ relate), or we change change the plate separation to actually change the capacitance. Changing one doesn't automatically change the other. In terms of the geometry, these are the only two things that matter. (We could include a dielectric between the plates, then there is more things that effect capacitance than just the geometry of the system).

Notice that by just having the equations there is no way for us to know any of this. Only when we know what the equations mean and what capacitance actually is physically can we make this sort of analysis of the equations in terms of what they tell us about the physical world.

$\endgroup$
  • $\begingroup$ There's also an implicit assumption that the parallel plates are very close together. If they're not, then the size and shape of the plates does matter as well. $\endgroup$ – Michael Seifert Dec 19 '18 at 15:44
  • $\begingroup$ @MichaelSeifert Yes that equation does assume "infinitely large and infinitesimally thin" plates. $\endgroup$ – Aaron Stevens Dec 19 '18 at 15:56
0
$\begingroup$

As you infer, the capacitance of an object determines that ratio of charge separation to the voltage associated with that separation: $$\frac{Q}{U}=C$$ using the same symbols as in the question. If $Q=0$, then $U=0$, but $C$ is non-zero.

The ability of a device to hold charge depends on the mechanical structure and geometry of the device, and the resulting potential difference between the locations of separated charges is proportional to the charge.

For two parallel plates, the ratio, geometrically, works out to $$C= \epsilon\frac{A}{d}. $$ Two conducting parallel plates, charged or not, have this capacitance. On the other hand, the capacitance of two concentric spherical shells is related not to an area, but to the product of the radii of the shells, $a$ and $b$, $ a < b $: $$C=4\pi\epsilon\frac{a b}{b-a}$$ If we expand the outer shell to infinity, and leave the inner shell at radius $a$, we get that a single spherical shell has a capacitance of $$C=4\pi\epsilon a.$$

So, the answer to your question is "maybe". For the concentric spherical shell, it doesn't depend on the area in the formula, but on the product of the radii, which has area units. But one might conceptually think of $4\pi a$ as the area divided by the radius.

But the geometric formula you give for the capacitance is only directly applicable to parallel-plate capacitors.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.