# Some questions about the capacitor

1. May 11, 2006

### x64bob

I've read about the capacitor on wikipedia but diden't catch some details; if the capacitor consists of 2 plates that can be charged and an insulator between them what connection do they have as i've read that the capacitor always is a a zero charge because of each plate always having the equal number of electrons and holes. can electrons pass trough the capacitor as they do trough a conductor(pic 1)? and if these plates's capacitans dont depend on connection between em could you use them separately like just having an - charged capacitor.

and would a scheme for stabilizing electrical current (filling out holes when there arent enough electrons in the flow and absorbing them when theres an excess look like this(pic 2)?

#### Attached Files:

File size:
68 KB
Views:
185
• ###### pic2.bmp
File size:
68 KB
Views:
198
2. May 11, 2006

### Staff: Mentor

The capacitor works because of the electric field that is created between the plates due to having more electrons on one plate compared to the other. Once in this static condition (with Q=CV, charge = capacitance * voltege), no currents flow anywhere in the circuit. The *net* charge of the two capacitor plates is zero, since you've basically "pumped" some of the electrons from one plate over onto the other to create the imbalance of electrons.

Capacitors are usually used as a storage element for charge (like in power supplies), or for their AC impedance characteristics. The AC current that flows "through" a capacitor is called the Displacement Current. It's called that because you have to displace electrons (pump them through the exterior circuit) from one plate to the other in order to change the voltage across the capacitor. As the voltage across the capacitor increases, the electric field E between the plates of the cap increases. This is the limiting factor in the operating voltage of most capacitors -- too high of a voltage will cause the dielectric between the plates to break down, and you will arc through the capacitor.

3. May 12, 2006

### x64bob

k, but i dont see why the electric field is needed here, or is it just something that happens if so you should be able to have a separate plate alone thats just charged with electrones and another with just holes?

in other words how do they depend of each other, why cant you just connect the electron chargable side and load it without charging the other?

Last edited: May 12, 2006
4. May 12, 2006

### Staff: Mentor

Keep in mind that the capacitor plates and external wires are made of conducting metal, not semiconductor material, so the concept of "holes" does not come into play. The mathematical concept of holes really only applies to semiconductor devices.

When a battery or other voltage source pumps electrons from the + plate to the - plate of a capacitor, the resulting imbalance of charge generates the electric field. You could also deposit some static charge (like in static electricity) on one isolated plate of a 2-plate capacitor not connected to anything, and yes, that will generate an electric field and a resulting voltage difference between the two plates. But that kind of arrangement is not very useful for anything in electronics. As I mentioned, caps are useful for storing charge in power supplies, and for their impedance characteristics in filters and related circuits.

5. May 12, 2006

### es

There is a "connection" between the plates but probably not in the sense that you are thinking (you know, like a wire soldered directly to both)

I am sure you will agree that an electric field extends outside of a charged plate and that there is an effect on a plate when it is inside an electric field.

So a capacitor does the obivous and takes plate 2 and places it in the electrical field of charged plate 1 causing an effect on plate 2. Thus plate 1 and 2 interact with each other through the elctric field. Stated another way, the plates are "connected" through the electric field.

6. May 12, 2006

### es

Oh ya, keep in mind that electrons and electric fields are different things. In general insulators are designed to stop the flow of the former, not the later.

7. May 13, 2006

### x64bob

ok, i do understand that there isen't any connection between the plates where electrons can pass and thats there is an electric field affecting the plates. but is the electric field only generated by the plate which get filled with electrons or do they both got field affecting eachother? also what happens when plate 2 gets affected by the electric field, i haven't read up on electric fields so does anyone knows a good source for that?

8. May 13, 2006

### 3trQN

Allways fun to watch these http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm [Broken]

Last edited by a moderator: May 2, 2017
9. May 15, 2006

### es

Yes. The plates effect eachother and in the ideal case they are symettrical. So it's your external influence that gives them the distinction of plate 1 and 2.

Note that for actual circuit implementations you may want to optimize the capacitor for this or that behavior and sometimes this optimization breaks the symettry. This is why sometimes when you open a device you see capacitors with + or - labels on the ends.

If you are interested in more information I also highly recommend the MIT video lecture series.

Last edited: May 15, 2006
10. May 19, 2006

### Mica

The formula to calculate the capacitance is C = (epsilon*A/d). Let say I have two plates of 1m x 1m, d = 0.03m and epsilon = 3.2 then C = 0.9 nF.
If I able to separate the plates (d increase) then my capacitance decreases. Assuming charges conservation, Q0 = V0 * C0 and Q0 = Vf* Cf, then I can rewrite the equations: Vf = V0 *C0/Cf. In this sense, I amplified my voltage. So more I increase d, more I can amplified my voltage. What is the limitation? I think we can not increase d to infinity. I understand that the RC decay time plays a role but if we able to increase d fast enough, we still have a huge amplification. In this example, if I able to increase d to 3m less then my RC decay time, I would have a factor of 100!! If I charge my capacitor to 1 V, the output will be 100 V. Then to a higher voltage, 1 kV input, the output would be 100 kV. So, what is the limitation in the real world? Do we definitively have these amplifications or is just mathematics?

Thanks,

Mica

11. May 19, 2006

### Staff: Mentor

The flaw in the large amplification part of your thought is the simplistic form of the equation for capacitance. This equation is only valid for a capacitor where the size of the plates is much larger than the separation distance:

$$C = \epsilon_0 \epsilon_R \frac{A}{d}$$

When the size of the plates is large compared to the separation distance, the fringe electric field at the edges of the plates does not contribute a significant portion of the capacitance. However, as you pull the plates apart (which takes work, but, because of the attractive force of the opposite charges on the two plates), the fringe field contributes to the overall capacitance in a larger and larger proportion. So when you get the plates way far apart, they look more like point charges out in space, and you would need to calculate the capacitance differently (quiz question -- what is the equation for the capacitance of two point charges separated by distance d?).

Keep on thinking about stuff like this. It's good to morph the situation some to see what happens. Good question, Mica.

12. May 19, 2006

### Mica

I'm thinking of some kind of dipole, but I can not connect the equations together. I don't kown.

So, what happens when I decrease the area of the plate? If I have two same parallele plates initially then, I slide one of them to decrease the length. Is it the same explaination as above?

Thanks for the quick response.

Mica

13. May 19, 2006

### Staff: Mentor

Pretty much. The simplified capacitance equation really only holds for when d is small, so there is not much contribution from fringing field. To the extent that there is significant electric field from one plate to the other that is not confined to the volume directly between the plates, the measured (or calculated with the full integral) capacitance will be bigger than the simple equation gives. For example, when you make a 1cmx1cm capacitor from two sides of copper clad PCB (0.062" thick typically, with $$\epsilon_R = 3$$ you get about twice the capacitance you would expect from the simple equation above.

14. May 19, 2006

### Staff: Mentor

BTW, The relative dielectric constant of the dielectrics used in bypass caps (like X7R, etc.) can be as high as 1000 to 10,000 ! In those cases, the fringe field out in the air where $$\epsilon_R = 1$$ is tiny.

15. May 19, 2006

### Mica

When you say small, what is the ratio for it? In the last case, d is constant.

What kind of material is with a dielectric high as 1000 to 10,000? The material with the highest dielectric constant is 10 that I know. I don't quite understand that the fringe field out in the aire is tiny? What do you mean by this?

So, we can not amplify the voltage with variable capacitor? Right?

Thanks,

Mica

16. May 19, 2006

### Staff: Mentor

As for the ratio, I'd guess L/d > 100:1 or so for air or low-value dielectrics, but that's just a SWAG. I understand that in the last example you are just shrinking one plate, while d remains constant. But sketch what happens to the fringe field to see what I mean when I say that shrinking one plate begins to invalidate the simple equation that you were using. Sketch it from the side view, with a top and bottom plate (each with length $$L = \sqrt{A}$$, separated by distance d. Draw the E field lines from the + to - plates. The E field lines are straight for the area between the plates that is near the center of the plates, and near the edges of the plates, the E field bows outward as it goes from the + plate to the - plate. At the very edge of the plates, there is even a little fringe field that bows way out to the side.

Now draw the case that you mentioned, where you shrink the top plate, say, by a factor of two horizontally. Now draw the new E-field lines. You see that there is a lot more fringe field bowing from the edges of the shorter top plate down to the wider bottom plate. Since a lot more of the E field is now not contained directly between the two plates, your original equation for C is definitely not valid.

First, you can look up typical dielectric materials used in caps to see what their dielectric constant is. I was amazed just like you when I saw the 1000-10,000 numbers. Wow. They must be some special materials, that's for sure. Just google X7R dielectric or something similar.

When I say that the external field will be tiny, it's because the field strength ratios with the dielectric constant. That means that as the dielectric constant of the material between the plates increases, the percentage of the electric field in the fringe air E-field lines versus the total decreases. The increasing of the dielectric constant between the plates effectively "focuses" more and more of the available E field in there, so less is out in the fringe field in the air. Think about the diagrams you've seen in your textbook that show how E field lines are "attracted" to a dielectric body that is placed in the air. It's the same kind of effect with magnetic field lines being bent toward ferrous metal objects placed in the air.

Actually, your original scenario works for the initial part of the increase in separation distance d. You will get an increas in V as you pull the plates apart, because Q=CV, and you are decreasing C while holding Q constant (assuming you've disconnected any external path between the plates). You just don't get a linear increase in V (or a linear decrease in C) as you keep pulling the plates apart. The change becomes less than linear with increasing d, about the time that d is not << the length of a side of the capacitor.

Make sense?

17. May 23, 2006

### Mica

Thanks for the detailed explaination. Does existed a formula or a text book that explained the limitation of d? What I mean is at some point (dmax) the V will decrease instead of increasing. I would have a formula or a graphic shows this behavior.

I saw a paper in MEMs that they varie the capacitor laterally, you can dowload from the web "Vibration-to-Electric Energy Conversion" from Scott Meninger.

Thanks,

Mica

Last edited: May 23, 2006
18. May 23, 2006

### Staff: Mentor

No, it will just stop increasing.

The capacitance is calculated by C = Q / V, and V is related to the electric field in the geometry of the capacitor arrangement by an integral. Take a look at any intro E&M textbook, and it will generally have a chapter or some sections on how to calculate capacitance for various geometries.

19. May 23, 2006

### es

That's an interesting idea. I usually thought of this capacitance as a nuisance.

Are there any typical applications for caps that use the PCB as the dielectric?

I suppose the obivious one is to measure $\epsilon_R$ directly for process control or something. Or since $$\epsilon_R$$ is typically well controlled then perhaps it could be used as a cheap/fast way to find board thickness, again for process control.

Last edited: May 23, 2006