Capacitance and Voltage

[Dr Gerrard]

So, apparently, when a capacitor is charging, the potential difference between the two plates increases, until eventually it stops plateaus off.

Also, C = Q/V, so here, as potential increases, capacitance decreases.

What is the difference between these two potentials?

Is this saying that the greater maximum potential difference between the plates is, the lower the capacitance, but still, as the capacitor is being charged, the potential increases?

The maximum potential difference can be found by calculating the maximum charge that the capacitor can hold, Q, and then determining the potential by KQ/r.

Then as the potential reaches this maximum value, the capacitor is charging. However, if this max potential was lower, then the capacitance would increase.

Is this right?

I don't know if that makes sense or not, but this doesn't make much sense to me anyways,

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[Chowdder]

Not quite follow your question....perhaps try and explain it a bit more.

I'll try and throw some explainations in hope that it might answer your question.

In reality, capacitance of a capacitor (the ability to sort charge) is not determined by Q or V. It is determined by relative static permeability, permeability of free space, distance separating the plates, and the area. Capacitance in a circuit is a constant. This is just like resistance is determined by the resistivity of the material rather than I and V. Just like you can still have resistance when there is no I or V, you can still have capacitance when there is no Q or V. The C=Q/V expression is similar to Ohm's law which describes a relation between Q and V that is dependent on a constant, C.

Also think of it like this, when you are charging a capacitor, you are increasing the voltage. At the same time you are also increasing the charge on the two separate plates. Increase in Q would cause an increase in V. This will be cancelled out leaving you with a constant capacitance.

You can work out the equation and see for yourself.
To charge a spherical capacitor, you have C=Q/V you have also mentioned that V=kQ/r, plug that into the equation and you get C=R/k, R do not change during charging and k is a constant. In parallel plates, the equation comes out a little different but the values of C at the end is still a constant.

Not sure if you're any clearer

 

[Dr Gerrard]

Sorry about not being so clear, let me try to explain myself again.

In EK, it says that the capacitance of a capacitor increases as voltage decreases. I am just trying to understand how this make sense.

Especially because as a capacitor is charging, the voltage between the plates increases. How does this increases voltage relate back to the total capacitance of the capacitor?

EDIT --- I realize now that capacitance is the same while it is charging, because as voltage increases, the charge also increases. Thus, capacitance is a quality of the capacitor.

Still though, why does EK automatically assume that as Voltage increases, capacitance decreases? What voltage are they talking about here?

I really hope that makes sense, but if not, please let me know.

 

[Chowdder]

I do believe they are referring to the potential different generated by the separation of charge between the plates of the capacitors. I haven't come across any question that ask about voltage dependent capacitors (it can get quite complicated and messy). I just stick with thinking capacitor is like resistors. Both are constants and both can be described with simple equations.