Concentric Sphere Discussions & Math

Updated: 5-20-11 & 8-13-09

With the outer sphere floating, it is just an equipotential surface around the inner sphere, and doesn't affect the capacitance of the inner sphere, or its breakdown voltage (ignoring irregularities due to the presence of the coil below). This can be easily verified by measuring the resonance frequency of the system, with the spheres insulated and interconnected. The combination would start as a small capacitance, but as soon as a spark puts the spheres in connection the effective load would be the outer sphere alone. In another post I said that a possible reason for improved performance would be the fast variations in the electric field around the outer sphere due to the sparking between the spheres. The rectification may have some effect too.

Thanks, I had seen your paper already. Does the equation in section VIII produce capacitance in terms of length? It is unclear to me.

The formulas apply to two spheres with radii a and b, with a distance c between their centers, c>(a+b). The formulas for one sphere inside the other are different. You have to compute first the terms below and then evaluate the series to find the three capacitance coefficients k11, k22, and k12. The actual capacitance -between- the spheres, assuming opposite charges on them, is given by (k11*k22-k12^2)/(k11+k22+2*k12). You see that the formulas are complicated, but trivial for a computer. The Inca program can do all the calculations. http://www.coe.ufrj.br/~acmq/programs Antonio Carlos M. de Queiroz.

*****Henry Cavendish (1731-1810) perfected the suggestions of Franklin and Priestley with his ingenious concentric sphere experiment in 1772. The inner of two concentric conducting spheres was first charged, then connected to the inner surface of the outer sphere. On then testing the inner sphere for charge, it was found to be uncharged. This showed that the exponent in the law of force differed by less than 0.02 from 2. Cavendish's work was nearly forgotten until resurrected by James Clark-Maxwell nearly a hundred years later. The experiment was redone and brought the limit even closer to zero. Modern experiments have refined the measurement even further. In modern terms, this is equivalent to measuring the "mass of the photon," which on theoretical grounds should be zero.I dogpiled 'concentric spheres' and here's some of the info I've uncovered:

## A Conducting Spherical Shell

Consider a conducting spherical shell with a net negative charge. At equilibrium the excess charge is uniformly distributed over the outside edge of the shell.

None appears on the inside edge because that would mean field lines would either pass through the shell or would converge at the center. Field is excluded from the bulk of the shell, and field lines would not converge at the center unless there was a charge there.

If the outer radius of the shell is R, the field is zero for r < R, and the field looks like that of a point charge (kQ/r

^{2}) for r > R.## Concentric Spheres

What happens with two concentric spheres, or with a point charge at the center of a spherical shell? The net field outside any of the objects is simply the vector sum of the fields from the different objects. The field inside any conductors is zero - charge on the conductor shifts to ensure this.

In the example above, a positive point charge Q is placed at the center of a conducting shell that has a net charge of -2Q. What is the field as a function of r, if the shell has inner radius R

_{1}and outer radius R_{2}?For r < R

_{1}the field comes only from the point charge. The charge on the sphere does not produce a field in this region, so E = kQ/r^{2}directed out from the center.For R

_{1}< r < R_{2}the field is zero because that's inside the conductor. For this to be true there must be a net -Q on the inside surface of the shell to stop all the field lines coming from the +Q charge at the center.For r > R

_{2}the field once again looks like the field from a point charge equal to the total charge in the system, which is +Q - 2Q = -Q. Therefore the field here is also given by E = kQ/r^{2}, but now is directed in toward the center.

## Spherical Capacitor

The capacitance for spherical or cylindrical conductors can be obtained by evaluating the voltage difference between the conductors for a given charge on each. By applying Gauss' law to an charged conducting sphere, the electric field outside it is found to be

The voltage between the spheres can be found by integrating the electric field along a radial line:

From the definition of capacitance, the capacitance is

## Isolated Sphere Capacitor?

An isolated charged conducting sphere has capacitance. Applications for such a capacitor may not be immediately evident, but it does illustrate that a charged sphere has stored some energy as a result of being charged. Taking the concentric sphere capacitance expression:

and taking the limits gives Further confirmation of this comes from examining the potential of a charged conducting sphere:

Q. Two charge concentric sphere are shown below. R1 = 10 mm. R2 = 15 mm. The two spheres comprise a spherical capacitor. What is it's capacitance?

A.3.34x10^{-12 }F

Definition : Consider a set of charges producing a certain electrostatic field in a certain region. Alternatively, we could think of the potential at any point due to these charges. All the points having the same value of potential form a three dimensional surface in space, called an equipotential surface. Usually the equipotential surfaces are plotted spaced in equal steps. That is the potential difference between any two neighboring equipotentials is the same. Equipotentials for a point charge are plotted to the left.

Consider a single point charge Q at origin. The potential due to it is V() = k·Q/rImagine a sphere of radius R about this charge. Since all points on this sphere have r = R, they all have the same value of the potential, k·Q/R -- so this sphere is an equipotential surface. It is clear that any other concentric sphere is also another equipotential.

In two dimensions, equipotential surfaces become equipotential lines.

Just as the force of gravity is a constant anywhere on a particular contour, the electric field has no component along an equipotential. Hence, when a charge is taken along an equipotential at constant speed, no work is done on it. Another way to express the same idea is to say that the work done in moving the test charge q

_{0 }from the point 1 to the point 2 without any acceleration is_{ }q_{0}(V_{2}-V_{1}) = 0, since V_{1}= V_{2}on the equipotential.

Information provided by: http://lecture.lite.msu.edu

Capacitance of an isolated sphere If a sphere of radius r is given a charge Q then the potential anywhere on it is given by :-

Thus, the capacitance of the sphere is

It should be noted that the capacitance of the sphere depends only on its radius and on the permittivity of the surrounding medium.

Capacitance of concentric shells.

Let a and b be the radii of the two concentric shells such that a<b. The inner shell is given a charge +Q while the outer shell is earthed.

Hence, charge -Q is induced on the inside of the outer sphere and +Q on its outside. This free charge on the outside of the outer sphere is neutralized due to the earth connection.

Since the potential (V_{A}) of the inner sphere is partly due to its own positive charge and partly due to the negative charge on the inside of the outer sphere, this potential is given by,

The potential difference V between the two spheres is therefore given by:

Thus the capacitance of this system is greater than that of a single sphere. Note once again that the capacitance of the capacitor depends only on the radii of the two spheres and on the permittivity of the medium, and not on the amount of charge stored or the potential difference between its two parts.## Capacitance of two concentric cylinders (e.g. coaxial cable)

C = 2 * pi * epsilon * length / ln( rOuter/rInner)

this assumeslength>>r## Capacitance of two concentric spheres

C = 4 * pi * epsilon * rInner * rOuter/ (rOuter - rInner)

asrOutergoes to infinity, the fractionrOuter/(rOuter-rInner)goes to one, leading to the following handy equation:## Capacitance of isolated sphere

C = 4*pi*epsilon*radius = approx 111.2 pF/meterthis equation is derived from the equation for two concentric (nested) spheres, and letting the radius of the outer sphere go to infinity.

## Estimating the magnitude of the Electric Field

## shamelessly copied from Jim Lux's website (which has lots more good stuff, I learn something new each time I visit): http://home.earthlink.net/~jimlux/

A lot of practical high voltage design requires knowing what the maximum E-field is, for insulation design, corona reduction, etc. The exact field can, of course, be calculated numerically by solving Laplace's equation over a suitable field with appropriate boundary conditions. As complicated and time consuming as this is, it is necessary when performance is critical, in integrated circuit design, designs for absolute minimum cost, and so forth. However, for more run of the mill experimentation and use, where a little overdesign can be tolerated, approximations to the field are just as useful.

The mean or average field is just the voltage difference divided by the distance between the conductors. For the proverbial infinite flat plates, this makes the calculation simple.

Emax = Eaverage = V / Distance## For two concentric cylinders (i.e. like coaxial cable) the maximum field is:

Emax = V / ( Rinner * LN( Router / Rinner))

where:

Rinneris radius of the inner electrode

Routeris the radius of the outer electrode

LN()is the log base e of the argument

Vis the voltage between the electrodes## For concentric spheres, using the same variables, the maximum field is:

Emax = V * Router / (Rinner *(Router - Rinner))## For two parallel cylinders of equal radius:

Emax = V * SQRT(D^2 - 4 * R^2)/ (2 * R*(D-R)*INVCOSH(D/(2*R)))

approx equal to:V / (2*R) * LN(D/R)if D>>R

where:

Dis distance between the centers of the conductors

Ris the radius of the conductors## For two spheres:

Emax = approximately V/S * F

where:

Sis spacing between spheres= D - 2*R

Fis a field enhancement factor =

F = (S/R+1) * sqrt( (S/R+1)^2+8)/4

For spheres, if S>>R thenEmax = approx V/ (2*R)## For other configurations:

Emax = Eaverage * F

For sphere/plane:F = 0.94*Spacing/Radius + 0.8

For cylinder/plane:F = 0.25 * Spacing / Radius + 1.05-20-11

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