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Capacitance of Concentric Cylinders
Monday, 15 October 2007

In this topic we will calculate the capacitance of a system of concentric cylindrical shells.

Figure 1 presents the geometry of this topic. Two conducting and concentric cylindrical shells, infinitely long, constitute a system that some capacitance, C.

concentric cylinders
Geometry of this topic. The cylindrical shells are infinitely long (though not drawn that way). The outer shell is of radius b while the inner shell is of radius a.

Begin by noting that we can only determine the capacitance per unit length, C/L, because the total capacitance is an unreal value. The definition of capacitance is,

Image

where Q is the charge of the system and V is its potential. From this expression it is seen that capacitance is the amount of charge that can be stored in a system by holding it at a potential, V. The charge is the amount that can be held separate, not just the total. For example, in a parallel plate capacitor the charge used in this is equal to the absolute value of the amount on just one of the plates (if we used the total charge then we would have zero).

To determine the capacitance of this system we will place some charge of the cylinders. Put +Q on the center cylinder and -Q on the outer. The net charge of the system remains zero. This charge will distribute itself over the surface of the cylinders. Our expression for the capacitance per unit length of this system is,

Image

where Q = Q is known (we put it there ourselves). It remains to determine the potential, V, that is maintained between the cylinders by the separation of this charge.

Since we know where all the charge is in this system it is possible to determine the electric field everywhere. Knowing the electric field, E, between the cylinders allows for the calculation of the potential through the relation,

Image

where this represents the potential between the end points of the line l. The line taken here will be along the radial coordinate connecting the cylinders. The system is symmetric and this connecting line between the cylinders represents the potential between them at all points.

The symmetry of the system is further exemplified by the electric field pattern of the inner cylinder shown in figure here. The infinitely long cylinder produces a uniform electric field along the r vector in the cylindrical coordinate system.

electric field of cylinder
The electric field produced by the inner cylinder of net charge +Q is entirely directed along the radial coordinate.

Returning to the problem of calculating the electric field, recall Gauss' law,

Image

where Qenc is the total charge enclosed by an area A.

Since we want to determine the electric field between the cylinders it is necessary to find a surface that is everywhere perpendicular to it (i.e. a surface with a normal that is parallel to the electric field). The dot product in this is non-zero in such a case. Figure here illustrates the surface that satisfies this requirement. The normal vector of the Gaussian surface, A, is everywhere parallel to the electric field vector.

gaussian surface between cylinders
The dashed circle is a Gaussian surface that will allow us to calculate the electric field between the cylinders.

In figure here only one example electric field vector has been drawn. This field is everywhere parallel to the radial coordinate vector. The charge of the outer cylinder does not contribute to the total charge enclosed by the surface. The enclosed charge is determined entirely from that of the inner cylinder. This cylindrical surface is three dimensional and represents another cylindrical shell. The entire charge of the innermost cylinder is enclosed by our surface, Qenc = +Q.

The differential area element, dA, can be rewritten in terms of this geometry. The total area of the Gaussian surface is given by the expression for the surface area of a cylinder of radius r. The surface is defined at a fixed radial position, so only the axial (z) and azimuthal (Φ) coordinates are necessary to compute the total area. This is shown below.

Returning to Gauss' law, let us solve each side separately,

Image

where the last step makes use of the fact that I am saying the length of the infinitely long cylinders can be written as L.

Equating these results leads to,

Image

where this is directed along the radial coordinate vector.

Returning back to this, the length element is now only the radial component, E · dl = Er dr,

Image

where it is very important to remember the reasoning behind the order of the limits in the integral.

Part of the definition of electric potential is that the potential at infinity is zero. When calculating the potential it is necessary to perform the line integration beginning at infinity (or just as far away as possible) and work your way back in. That is why the integration limits proceed from the outermost point, b, and end at the innermost point, a.

The potential is negative, which presents a problem for us because capacitance is positive definite. The following identity is useful, ln(α/β) = -ln(β/α). The potential between the cylinders, after we place equal and opposite amounts of charge on them, is,

Image

and we can now return to this to calculate the capacitance per unit length of this system.

Image

where the capacitance per unit length is not a function of either the charge on the system or L itself. The only parameters that matter are the radii of the cylindrical shells. This should be expected because capacitance is a feature of a system's geometry and does not depend on applied charges or potentials.

Comments
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Justin A. - What if the outer cylinder is | 2008-01-14 19:39:19
Hello David, Thanks for the great write-up.
It brings me to a question that's bothered me for a while now. How would you go about evaluating this for an outer concentric cylinder infinitely far away? I assume that this particular write-up doesn't apply since you can no longer say that L >> b. I suppose the question reduces to: how much charge is on a cylinder in a vacuum?

Thanks again.
Justin
David Pace - Outer Cylinder Infinitely Far http://www.davidpace.com | 2008-01-14 20:02:42
Hi Justin,

That is an interesting question and something that is good for further developing a conceptual understanding. The first issue is that the charge on the cylinder can be anything. For capacitance, the charge on the object is free charge placed on it purposely, so the position of the second cylinder does not change that. In reality, there is a limited amount of charge that can be placed on any object, but determining that value is a separate issue.

Recall that capacitance is a geometric property of a system. Making one of the objects distinctly non-geometric, by way of making it infinitely large, prevents us from applying the method discussed in this entry. The outer cylinder has an infinite surface area, therefore we cannot place the same amount of charge on both cylinders. Trying to place -Q on the outer shell results in a non-uniform distribution (if it had a uniform charge density then the total charge would be infinite).

Hope this is helpful,
David
Justin A. | 2008-01-15 11:46:59
Hi David,

Good stuff. Thanks for the quick and thorough response. As is usually the case though, it brings up a slew of new questions.

Regarding the limited charge on an object, is that statement referencing the "work function"? I'm hazy on it all, but as I understand it-- it is a material & geometry property that limits how many electrons can be "bounded" on the surface of an object before they start ejecting from the local pressure of repulsion from neighboring electrons.

About charge, I guess I should have asked about, "the capacitance of a cylinder in a vacuum".

Since we can't evaluate the total capacitance of an infinitely large cylinder, can we evaluate the capacitance per length?

The only comparable calculation I've seen is where the capacitance of a single sphere in a vacuum is calculated. There, you use the concentric sphere capacitance and take the limit as the outer sphere has a radius of infinity. I guess I was thinking that something similar might apply, but I don't know. After doing some more searching, I still can't find where anyone addresses the capacitance of a single cylinder (infinitely long or not). It's probably very naive, but I would have guessed that the cylindrical geometry would lend itself to a direct analytical expression. Any thoughts?

Thanks again,
Justin
David Pace - Taking Limit to Infinity http://www.davidpace.com | 2008-01-16 15:24:24
Hi Justin,

I think you are correct about the issues related to limits on the amount of charge that can be placed on an object. The repulsive Coulomb forces eventually grow too large and electrons begin streaming off of the surface.

Concerning the capacitance of a cylinder in vacuum, it sounds like the limit as the outer radius goes to infinity is the way to go. You can take the result from this article and take the limit as b goes to infinity. The result here is already per unit length. I have not worked this out, but thanks for the idea because it might be worth adding it at some point in the future.

- David
Kwadwo | 2008-03-27 09:50:56
Hi,

How would go about the derivation for parallel cylinders of equal radius? Also,assuming there is a dielectric material e.g.ice between the cylinders,how does that influence the electric field generated?

Pace Replies:
Are you asking about cylinders that are next to each other, but not concentric? The procedure is the same as given in this topic. I can't explain the differences very well in this reply, but maybe I can write up the new problem (sorry, this will not help you because it will be a long time before I do that).

A dielectric material between the cylinders will increase the electric field and, therefore, also the capacitance. This is assuming that the dimensions are not changed. In such a case, the increased electric field is due to an increased potential. Very large capacitors (large in terms of capacitance, not size) are made by placing dielectrics between plates.

Thank you for your comment and questions.
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Copyright (C) 2007 Alain Georgette / Copyright (C) 2006 Frantisek Hliva. All rights reserved.

Last Updated ( Friday, 15 February 2008 )
 
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