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Physics/How to calculate the capacity of a metal plate

Question
QUESTION: Dear Sir,

I would like to evaluate the capacity “C”, of a metal plate, with the dimensions of :  axbxc, how can I find the capacity of a charged cube or prism, made of metal, with a certain length, width and thickness, it may be an approximate way of solution, what would you suggest ?

- I would like to suggest a way :
According the Gauss law, the electric field is the function of the charge and the charge is the function of the surface area, so q = f(A), the bigger the area the more charge accumulated on the surface. In my solution, I calculate the whole area of the prism :

A = 2 x ((axb)+(bxc)+(axc)), and equalize the surface area to a hypothetical sphere with the same area, so A = 4 x pi x r^2, by this way I get the radius of an equivalent sphere : r = (A/(4xpi))^(0.5), later, I calculate the C value, in a simple known way as,
C = 4 x pi x epsilon-zero x r, does this way make any sense ?

Yours Sincerely
Birol

ANSWER: It makes sense as an approximation, but in a practical sense it's not the greatest because the sharp edges will have very high electric fields.  The plate will also have to have small dimensions compared to anything it is around.  There is no simple solutions to this problem, it requires detailed knowledge of the sharpness of the edges.  I'm not sure what possible practical purpose the capacitance of a single plate has, it's easier just to measure it in-situ.

---------- FOLLOW-UP ----------

How can I measure the capacity "C" of such an geometric object, is there an easy way, or just over the voltage V, and i (current of the load R) which gives the XC value, and while the frequency is known, C could be evaluated in my opionen, am I right here ?

Yours Sincerely
Birol

You can measure the capacitance with a proper capacitance meter. The difficulty lies in finding a place to pick for the ground lead.  It has to go to "infinity," which could be the ground of the room that you are in.  These meters can be somewhat expensive, for high-fidelity, but you're not talking about 8 decimal places or anything.  A cheap one works fine.  The capacitance will be very small, in the pF or maybe up into the nF region, and I'm still not sure what the practical application of such a tiny capacitor is.

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