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QUESTION: Hello Steve,

I hope you are okay !

I have a problem for a while and trying to find a solution. Assume a capacitor (good quality) as given in the attachment file. I have one of these capacitors with a C value of 1 micro Farad.

I also check the chapter 26,Capacitance, in the book of : "Fundamentals of Physics", I was able to find the C value for a parallel plate capacitor, cylindrical capacitor and spherical capacitor by using the Gauss law !

Hmmm, how can I find the C value of a metal plate with a surface of a x b, and thickness (c) you may ignore it.....I mean a metal plate can hold charge, so, it must have a C value, how can we find it, or let me ask you in this way, what is the C value of each electrode (A or B, anyway they will have the same value) in the given 1 micro F capacitor ?

Is this also 1 micro Farad or much less than this value, how can we apply the Gauss law, to derive an equation which may give the C value for a parallel plate capacitor ?

Many thanks,

Birol

ANSWER: What you have here is a parallel plate capacitor. The layers each act like a parallel layer, so you just add in the capacitance of all of them. Because you can alternate the plates which go to each contact and each plate (due to its finite thickness) can have charge on both surfaces, you just count the number of gaps. So use the formula for the capacitance of a parallel plate capacitor (epsilon_0 * Area/gap)*(dielectric constant of material in the gap)*(number of gaps between plates). That last one is to add all the capacitors in parallel. They are all in parallel, sharing charge, so the capacitance just adds up...hence the last factor in the formula.

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QUESTION: Hello Steve,

Thank your for your try.

Can you say that one electrode of the capacitor may have a capacity of 0,50 micro Farad ? (As you know there are 2 electordes in a cap plus dielectric medium). I understand what you mean, you see it as multi layered cap, with many parallel layers, however, there are still 2 electrodes, it is irrelevant how many layers they contain......

Let me change the question in this way, you have a parallel plate capacitor, each plate has an area of 0,04 m^2 (20 cm x 20 cm), okay, the distance between the plates is 0,001 m, assume we have silicon between the plates with a relative permittivity of 11,68, so, we would calculate the C value, as 4,13 exp(-09) Farad. How would you evaluate the C value for each plate ?

Or, you may also assume that we have just one copper plate (sheet), of the same dimension : 20cm x 20cm, how would you find the capacitance (at the surface) of this copper plate ?

It must have a value, while it can hold charge, am I right ?

An isolated metal sphere would have a capacity of C = 4x Pi x epsilon_0 x R), I want you to make the same calculations for a metal plate with dimensions of a x b x c (c may be very tiny)

Many thanks

Birol

ANSWER: It is absolutely NOT irrelevant!!! The surface area is relevant, not the number of electrodes!!!

Basically, you diverged into where your questions show that you don't quite get the concept of capacitance yet. Let me try to set you straight, but this is not the forum for a whole physics class.

For each plate here is different than the question you ask at the end. There is not capacitance for each plate here, there's a capacitance for the pairs of plates. Multiply by the number of pairs and you get the total capacitance. You got the capacitance right for one parallel plate capacitor (I have a different number in the 3rd decimal place, I'll assume it was rounding off or something). But if that's the area of each plate then you have to multiply by the number of layers. If it's the area of all the plates added together, then it's correct and for "each pair" you have to divide by the number of plates.

A sphere still has to have a defined zero potential. If you have a sphere, you call it infinity and integrate against its electric field to find the voltage difference at its surface. If it's a plate, you have no such nice and easy definition. You cannot easily define the capacitance for any geometry except a sphere with a single electrode, and a set of plates makes the question itself meaningless. Sorry. You can't find its electric field relative to infinity without doing really nasty integration and assuming a thickness of plate.

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QUESTION: Hello Steve,

Thank you. As I guessed, it seems not easy to calculate the E Field, V and so the C of a single plate.

Why did I ask you this question.....I did an experiment, I used a capacitor of 1 micro F capacity, I pulsed just one electrode with AC, and I was able to observe the signal on the second plate (electrode) of the capacitor, the current was low, instead 30 mA, we measured 2-4 mA, I am trying to understand the reason, if C is low (for the initial electrode) Xc would be high, and this would restrict the current.....What do you think, can this be the reason ?

Birol

It makes no sense to calculate the capacitance on a single plate when you have two plates. Those are totally different geometries with different capacitances. The value of Xc will depend on the frequency. I don't understand your question, or why you were expecting 30 mA. Surely you didn't get 30 mA onto one capacitor plate and not 30 mA off of the other plate...charge doesn't build up in a capacitor like that (with one sign larger than the other).

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