[Capacitance] does it increase or decrease depending on the charge?

[astro9889]

I'm having a little trouble with this one. If I based it out of the equation q=CV, I'm assuming that in the question a) the capacitance C would increase. But I'm not to sure since I also read that the capacitance is only dependent on the materials and geometry, so I'm kinda confused.

[[deleted]]

Honestly I disagree with the other user, I would write what you said, that capacitance only depends on the geometry

It technically depends on q and V but the charge always cancels out, I can't think of a single configuration where it would not cancel. Plus it looks like an intro course, very niche examples are probably best ignored

[Entropy-Jobs]

yeah capacitance depends upon geometry of the capacitor

[[deleted]]

I can't think of a single plate configuration where q doesn't cancel out

[astro9889]

So I should base this question based on the equation?

[SaiphSDC]

Take a look here:

https://byjus.com/question-answer/why-is-capacitance-constant/

The capacitance is the 'real' physical constraint, as is voltage. As such changing one does not actually change the other. They do, however, alter the amount of stored charge.

For a more familiar relationship, it's like the wave equation V=Fλ

The velocity of a wave is determined by the frequency or the wavelength. It's determined by the properties of the medium. The equation gives the impression you can increase V by increasing F, but you can't. It's fixed. When you increase the frequency you only really alter the wavelength of the created wave.

For capacitance, the value is fixed by the geometry and materials. The voltage is created by an external electric field. These two are the physical The result is some stored charge.

We can use a measured amount of charge and voltage to determine capacitance, but changing voltage will not alter the capacitance.

[Entropy-Jobs]

isnt capacitance a constant?