Capacitance of a cylindric capacitor (only a piece)

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To calculate the capacitance of a 30-degree segment of a cylindrical capacitor, the traditional cylindrical capacitor formula may not apply due to the loss of symmetry. If the field gap is small compared to the capacitor's dimensions, a parallel plate approximation or a modified cylindrical capacitor approach can be used, dividing the capacitance by 12. For more complex scenarios, the method of moments can provide a detailed solution, accounting for fringe fields, as outlined in Harrington's or Gibson's books. This method is more intricate than the simpler approximations suggested. Understanding these approaches is crucial for accurately determining the capacitance in non-symmetric configurations.
henry7512
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Hi..
I´ve been trying to solve this problem for a couple of weeks and I need fresh ideas...it´s a 30 degree piece of a 10 cm high cylindric capacitor..how can I calculate its capacitance if it stopped to be symmetric?From my point of view the field between the 2 plates shouldn´t be calculated the same way we use for the cylindric capacitor..anyway..if the field can´t be calculated that way then I think it gets to complicated to do it...
Thanks for any idea (as long as it´s good)
 
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This is a good question. If the field gap is very small relative to the other dimensions of the capacitor plate, the fringe field is small, and you can use a parallel plate capacitor approximation, or use the cylindrical capacitor solution for a 360 degree capacitor, and divide the capacitance by 12. As a last resort, look at the problems at the end of chapter 4 in Smythe "Static and Dynamic Electrricity" 3rd Ed. for problems that calculate the edge correction for bulging fields.
 
thanks Bob..that was very kind of you
 
If you want to go all out you can also try a method of moments solution. Harrington's book or Gibson's book both give good details on how to solve electromagnetic and electrostatic problems using the method of moments. They both have the problem of a charged plate. This solution will take into effect things like fringe fields and such but will be much more involved than the approximations that Bob has suggested.
 
It may be shown from the equations of electromagnetism, by James Clerk Maxwell in the 1860’s, that the speed of light in the vacuum of free space is related to electric permittivity (ϵ) and magnetic permeability (μ) by the equation: c=1/√( μ ϵ ) . This value is a constant for the vacuum of free space and is independent of the motion of the observer. It was this fact, in part, that led Albert Einstein to Special Relativity.
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