Capacitance of parallet plate with dielectric spheres in between

AI Thread Summary
The discussion focuses on calculating the capacitance of a parallel plate system with dielectric spheres of varying radii and dielectric strengths between the plates. The capacitance can be derived using the formula C = permittivity * dielectric * area / distance, with the understanding that the arrangement of spheres introduces complexity due to varying distances at different points. It is suggested that the integration of capacitance values is necessary because the effective dielectric distance changes along the surface of the spheres. A practical approach is recommended, using square dielectrics arranged in series and parallel configurations to simplify measurements. Overall, the conversation highlights the challenges of theoretical calculations versus practical measurements in capacitance determination.
amisha
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I want to calculate the capacitance of a system consisting a parallel plate and dielectric spheres of different radius and dielectric strength in between the plates.
please give some mathematical derivation.
I really need to know it,my career depends on it.
Please help me.
:)
thanks & regards
ruhi
 
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capacitance in parallel is same as resistor in series and vice versa.
c = permittivity*dielectric*area/distance

If you have a sphere, it will be an integration of series of capacitance depending on the diameter of the sphere at different location. At the tip of the sphere, the dielectric distance/space is less compared to the middle of the sphere. For example if you had a circle, the diameter maybe say 1m, but it is not 1m x 1m. Instead it is 1m diameter but as you cut and move along the circle, it becomes 0.99m,0.98m,0.979999m etc.

For start, i suggest using a square dielectric. Take 3 squares, with 2 of them being the same length but 1/2 the length of the first one. Put the 2 smaller square in series (next to each other). Stack the bigger square on top of them. The circuit will look just like 2 resistor in series parallel with the other.
 
This might be a situation where doing a measurement is much easier than doing the calculation.
 
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