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University problem. Help please?
Given a horizontal uniform electric field E, calculate the electric flux that enters the left hand side of a cone with base radius R and height h.
Note: Just to clarify, the cone is pointed up (or down, doesn't matter), and the electric field goes through it from the side. Something like:
-->.../\
-->../ .\
--> /__\
Actually, I was thinking just now...
Since all flux that goes in must come out (by law). We could redraw the cone as half a cone on the side the flux is going in, and just a flat surface on the other side.
Thus, the amount going into the cone should be equal to the amount going out on the other, flat, side (which is just a triangle!).
Does that work? Can I just calculate the flux going through a triangle of area = area of cone projected on the y-z plane?
2 Answers
- helloLv 68 years agoFavorite Answer
Isn't the area just a triangle of base 2R and height h so area = 1/2 2R h = Rh
flux = field x area = E x r h = ERh
- 8 years ago
My E&M skills are somewhat rusty, but I think you should be able to calculate the surface area of the cone (exclude the bottom!) then divide that by 2. This should give you the amount of surface area exposed to the flux. From here, be sure to calculate the angle between your vector field and the side of the cone. The rest should be "plug and chug" using a Gaussian Surface equation.
Like I said, I'm a little rusty on my E&M so, this may or may not work. And, I may have overstated the obvious, if so, I apologize.
