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Re: Re: [TCML] Calculating the C of multiple Spheres for a top load ?



On 27/2/2012 22:49, jhowson4@xxxxxxxxxxx wrote:
Antonio,

What did you end up doing for your 4 sphere approximation ,

I was attempting to use image charges, (3 per sphere,each on a line to the center of another sphere located at some radius less than the sphere radius.) to figure it out but with my senior project being due last week I did not have the time to look into it much aside from the casual thought.
I used a quite crude "method of moments" technique:
1) Assemble the 4x4 matrix [P] relating charges and potentials, V=[P]Q. Assuming unity charges, the diagonal elements are the inverses of the capacitances of the isolated spheres (1/(4*PI*e0*a), a=radius), and the others the potentials generated at the center at a sphere by an unity charge at the center of the other (1/(4*PI*e0*d), d=distance). All these values are approximated, exact only if the spheres are far apart, but the error is small. 2) Invert the matrix, obtaining the capacitance matrix [C]=[P]^(-1) and add all the elements. This is the desired capacitance. For 4 spheres centered at the corners at the square with side c, each with radius a, I get:
C = 4*PI*e0*4*sqrt(2)*a*c/((2*a+c)*sqrt(2)+a)
I can't test how precise this is, but I tried the same method for axially simmetrical configurations, that I can verify with the Inca program, and the values agree within 3%, even when the spheres are touching, when the error is maximum, and tend to be exact when they are at some distance.

About other problems involving capacitances of spheres. The capacitance of one sphere of radius a is just 4*PI*e0*a, with e0=8.85e-12. For two spheres in contact, exact closed expressions are known too. For two separated spheres the capacitance matrix can be found by infinite series. I could not find published solutions for cases of more than two spheres, but It seems that the method of multiple images can be applied.

Antonio Carlos M. de Queiroz

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