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Re: [Phys-L] Charge division on Spheres.



I haven’t confirmed anything, except that geometry matters in any real experiment.

For two concentric spheres, C goes like R1 R2 / (R2 - R1). If *and only if* R2 is infinite do we get C proportional to radius R1.

For two coaxial cylinders (with no endcaps because infinitely long) C goes like L / ln(R2/R1). This does *not* scale with area (or any other simple geometrical factor).

For two parallel plates (which is my model of the endcaps) C goes like A / d. This doesn’t scale with area A alone because it also depends on the gap distance d.

I see no case (either theoretically or experimentally) where the answer scales with surface area alone. Also, to get the isolated sphere approximation, the surrounding “outer” metal bucket has to very far away from all points on the surface of your object (the soup can in your experiment). -Carl


On May 5, 2017, at 12:52 PM, brian whatcott <betwys1@sbcglobal.net> wrote:

Carl, I was hoping to clarify the law of capacitance of spheres when you reminded me the capacitance goes as radius and I was supposing C goes with (surface area) in error, when you said Q1/R1 = Q2/R2.

And playfully, I showed a correspondence in close quarters, that associates capacitance with surface area - and in response, you provide a more careful model of cans in an enclosing concentric cylinder with end caps which seems to confirm my mistake: that C ~~ Surface area. I am CERTAINLY missing something here. I expected you to say simply: the capacitance relation changes from surface area of a sphere at small separation from the grounding enclosure to the sphere's radius at very big separation from the grounding enclosure. Is that it??

Brian W

On 5/5/2017 6:44 AM, Carl Mungan wrote:
Photo of largest can in place before connections and lid placement
http://s880.photobucket.com/user/betwys/media/Capacity%20of%20%20Right%20Cylinders/IMG_1933.jpg.html <http://s880.photobucket.com/user/betwys/media/Capacity%20of%20%20Right%20Cylinders/IMG_1933.jpg.html>

I’m afraid that’s a very tough geometry. The rims of the two ends of the can are quite close to the outer pot and will dominate the capacitance in a tricky fashion. If we try my simple model:

C = 2 pi epsilon0 LENGTH / ln(0.23/DIAM) + pi epsilon0 DIAM^2 / d

where d = diameter of steel pot - LENGTH

then you get answers that follow the linear trend of your measured values but are only about 65% of the measured values. Again, the small spacing between the rims and the pot is the likely culprit.

Do you have a deeper pot so you can put the can vertically instead of horizontally? Then I would expect much better agreement with my model above, replacing “diameter of steel pot” with “distance from bottom to top cover of pot”. You should be able to try this at least with your small pots.

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Carl E. Mungan, Professor of Physics 410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu http://usna.edu/Users/physics/mungan/

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Carl E. Mungan, Professor of Physics 410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363
mailto:mungan@usna.edu http://usna.edu/Users/physics/mungan/