- #1

- 18

- 0

I cant get through the math part.can someone write and simplyfy it?

because a disk is an ellipsoid where semi major axis approaches zero and semi minor axis approaches R then c approes iR !!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter sentinel
- Start date

- #1

- 18

- 0

I cant get through the math part.can someone write and simplyfy it?

because a disk is an ellipsoid where semi major axis approaches zero and semi minor axis approaches R then c approes iR !!

- #2

- 18

- 0

- #3

- 104

- 0

Not sure what alternate method you are suggesting here.

I thought you could only get the field on the symmetry axis without poissons?

- #4

- 18

- 0

so if we have a conducting ellipsoid,it means that we we have an ellipsoid at a CONSTANT potential>>so we simply forget about the ellipsoid and put a rod instead.in analogy with what we prooved we can simply find out a rod of what lenght and charge per lenght will do the job.

so we have the potential of an ellipsoid in space!!(we use the potential of rod except that we right its landa and lenght(L) in terms of V of the conductor and its sami major axis).

now a disk is an ellipsoid really.(in the limit that its z lenght is zero.)

now we just have to take a limit of what we find for a rod to get the potential of disk in space.but it gets TRICKY!mathematically!

- #5

- 104

- 0

- #6

- 18

- 0

but we have the potential.we just need to plug in the according prameters!!which is much simpler then doing the BRUTE FORCE integral which cant be done!

Share:

- Replies
- 3

- Views
- 4K

- Replies
- 17

- Views
- 15K