Electric flux

  • Thread starter Will
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  • #1
Will
This is a proof type problem from our E&M class. We are given a point charge Q, let say that it is at the origin. We then have a disk parallel to the xy plane, with center a distance b from the origin( on the z axis) & radius R. We are to prove that if 1/4 of the flux passes thru the disk, prove that R= [(3)^.5]*b.
I assume that I have to use a spherical gaussian surface with 1/4 of its area corresponding to the size of the disk. Thats where I am confused. Do I make the top of the sphere just touch the bottom of the disk and have radius b, or do I construct it over the top of the disk, like a dome? And how do I acount for the fact that the electric feild lines wont be perpendicular to the surface of the disk because it is not round? Will this problem involve calculus, or is more of a geometry type problem?
 

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Hi Will,
I think it's a purely geometrical problem. If you construct a spherical dome over the top of the disc, then the dome's surface should equal 1/4 of the full sphere's.
 
  • #3
Will
Originally posted by arcnets
If you construct a spherical dome over the top of the disc, then the dome's surface should equal 1/4 of the full sphere's.

Thats what I concluded myself. So how do I solve for the radius of the disk in terms of its distance from Q? I was trying to reveiw double integrals in polar coordinates fo SA, does it need to be that complex? Am I missing something simple here?
 
  • #4
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Originally posted by Will
does it need to be that complex?
No. You could use the formula S = 2[pi]*r*h, where
S = surface area of dome,
r = radius of sphere,
h = height of dome.
Just find out how to express r and h in terms of R and b (which is easy), and there you are. OK?
 
  • #5
Will
Originally posted by arcnets
You could use the formula S = 2[pi]*r*h
So if one hase a sphere inside a cylindrical shell with equal radius and height 2r, then for any horizantal slice, parts of shell and sphere will have the same SA? Cant believe I did two years calc. and dont know that! Even looked thru a bunch of my old texts and did not see a formula like that written anywhere. Makes sense visually, and the problem became much simpler. thanks
But if I did do all that double integral-ly type stuff, I should get the same answer right?
 
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  • #6
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Originally posted by Will
So if one hase a sphere inside a cylindrical shell [...]
Yes. Geographers use this to construct world maps, see here:
http://mathworld.wolfram.com/CylindricalEqual-AreaProjection.html

There's a lot more formulae like this about spheres. Saves you a lot of time if you own some proper mathematical handbooks.
But if I did do all that double integral-ly type stuff, I should get the same answer right?
I'm sure you will.
 
  • #7
Will
Well, I don't feel so bad about not knowing this. I went to my old teacher for DFQ's and even he was scratching his head saying he didnt think so! So where can I get a handbook with a extensive collection of geometry stuff than my textbook? I head Shaums outlines are good.
 

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