# I don't understand capacitance

1. Mar 8, 2013

### Woopydalan

Hello,

So I am told that capacitance is a measure of how much charge can be stored in a capacitor. However, I don't have an intuitive understanding for why it is a relationship between charge and voltage. Can anyone help me in understanding it?

I'm imagining one capacitor attached to a battery. Electrons on one side of the plate leave that plate, go through the battery, and go over to the other side. If I increase the voltage of the battery, I would think even more electrons would leave the plate, yet the same amount always goes back to the other side. That ratio of charges to applied voltage is the same.

If I make the capacitor larger, is it the case that for a given amount of voltage, the same amount of electrons would leave one plate and go to the other side?

Last edited: Mar 8, 2013
2. Mar 8, 2013

### Drakkith

Staff Emeritus
When you increase the voltage more charges do leave the plate.

No. A larger capacitor loses more charges for the same applied voltage than a smaller capacitor does. This is why they hold more energy.

Capacitance is given by the equation C=q/V, where C is capacitance, q is the charges on the plates, and V is the voltage between the plates. Notice that this means that capacitance is a ratio of charges to voltage. A higher capacitance means that q is larger while V stays the same, or that V is smaller while q stays the same.

Also, a parallel plate capacitor's capacitance can be found by: C=εr ε0 A/d.

εr is the dielectric constant of the capacitor
ε0 is the electric constant
d is the separation between the plates
A is the area of the plates

Capacitance is proportional to the area of the plates since A is in the numerator. Doubling A will double your capacitance. If we double capacitance by increasing the area, but keep the voltage applied the same, then by the first equation we must increase the number of charges (q) moved.

3. Mar 8, 2013

### Woopydalan

Can you explain why the voltage is not the same for two capacitors in series? from a conceptual point of view, I understand mathematically why, but not conceptually.

4. Mar 8, 2013

### Drakkith

Staff Emeritus
5. Mar 8, 2013

6. Mar 9, 2013

### VCortex

I believe it's because capacitors connected in series effectively cause the sum of the "gap" between the capacitor plates over the whole of the circuit to be larger. I.E, the total gap in a series circuit is G1+G2+G3 etc for each capacitor added.

7. Mar 9, 2013

### Woopydalan

Also, I don't understand the connection between the capacitance of a lone object versus the parallel plate capacitors in a circuit.

Something like coaxial cables that are charged. If neither are connected to a battery, how do I know the capacitance of the cables?

This webpage derives the equation, but I don't understand why I only care about finding the electric field of the inner cable?

http://faculty.polytechnic.org/phys...._capacitors/2._pdf's/capac_of_coax_cable.pdf

Last edited: Mar 9, 2013
8. Mar 9, 2013

### Ratch

Woopydalan,

Which voltage? The voltage across each capacitor? If the caps have the same value, then the sum of the caps have to equal the voltage source.

What kind of lone object. Is not a single cap a lone object? What kind of connection?

The line integral of the electric field gives you the voltage. The field is between the inner and outer parts of the cable.

Ratch

9. Mar 10, 2013

### Woopydalan

Why is it that when you put a dielectric between the capacitors, they are modeled as being two separate plate capacitors? One with each plate and one part of the dielectric. I am not following the logic behind that.

10. Mar 10, 2013

### Ratch

Woopydalan,

If you put nothing but vacuum between the plates, then they are vacuum capacitors. If air is between the plates, then they are air capacitors. Mica makes a mica capacitor. Vacuum has the poorest dielectric constant, air is a little better, and mica the the best of the three I listed.

Ratch

11. Mar 10, 2013

### Woopydalan

Yes, but why is it that they are modeled as two separate capacitors?

12. Mar 10, 2013

### Ratch

Woopydalan,

Where? Show me.

Ratch

13. Mar 10, 2013

### Woopydalan

When it says that energy is stored in a capacitor, is that energy just the charges that are residing on the capacitor, or is the energy being held within the electric field between the two plates. If the latter, how is energy stored in an electric field if no charge is present?

Also, was that Ratch guy a crackpot or something? Looks like someone banned him

14. Mar 10, 2013

### Drakkith

Staff Emeritus
The energy is stored in a capacitor as charges in an electric field. If you short the two terminals of a capacitor together work will be performed on the charges by the electric field and current will flow. That's where energy comes from. The ability to perform work.

Note that you cannot separate charges and the EM field. Even when neutral there is a field present, it's just not capable of performing any work.

15. Mar 10, 2013

### Emilyjoint

Even when neutral there is a field present, it's just not capable of performing any work.
This is an interesting statement, can you explain/give examples of what it means?

16. Mar 10, 2013

### Drakkith

Staff Emeritus
I use the basic definition of a field from wiki:

While the field is a "construct" of ours using math, it represents what we think is a "real" field. We think this for a variety of reasons, one of the main ones is that energy is transferred through fields. We don't like to think that energy just disappears until it is transferred to another object so we say it has been transferred to the field.

When I say that the field is still there, but it can't do any work, I mean that the field, as per the description above remains, only the values change. Now that I think about it I am not 100% sure this is a correct view, so don't try to use this on any homework questions or anything.

17. Mar 10, 2013

### Emilyjoint

I do not recognise your wiki definition of a field. There is no reference to energy or energy transfer and it is not clear what is meant by classical fields or quantum fields.
If you are not sure of your facts or cannot back them up with recognised standard text books then you should think twice about what you present.
There is a great risk that students will be confused by opinions rather than knowledge.

18. Mar 10, 2013

### Drakkith

Staff Emeritus
My apologies I forgot to include the link to the article that touches each of those topics.
http://en.wikipedia.org/wiki/Field_(physics)

A reasonable objection. But I find that, in general, my opinions are usually accurate. Not always, but usually. If that changes then I will stop giving opinions.

19. Mar 10, 2013

### Woopydalan

so how is it that the energy is being stored within the confines of the two capacitor plates? For the electric potential energy to be present, doesn't there need to be actual charges that are present within the electric field? Or is it that the charges that are bounded to the plate capacitors are within the field, and that is the electric potential energy?

20. Mar 10, 2013

### Drakkith

Staff Emeritus
Let's be clear. When a capacitor has a voltage applied to it and charges move from one plate to the other, we say that an electric field is developed between the plates. If you had a vacuum between the plates there would be no charges "within" this field. But that's okay. The field is still there and the energy is stored in the separation of the charges.