electric flow


by ori
Tags: electric, flow
ori
ori is offline
#1
Oct30-04, 03:19 PM
P: 30
http://t2.technion.ac.il/~snoop/Q.gif
Q is at one point
R is the radius of this cylinder, it's height is 2h
the cylinder is without the bases.
how can i calculate the electric flow through it?
the final answer is Q/[epsilon0*sqrt(1+R^2/h^2)]
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Tide
Tide is offline
#2
Oct30-04, 03:30 PM
Sci Advisor
HW Helper
P: 3,149
Why don't you try calculating the flux through the flat top and bottom of the cylinder? Then you could subtract it from the total flux given by Gauss' Law.
ori
ori is offline
#3
Oct30-04, 03:38 PM
P: 30
Quote Quote by ori
http://t2.technion.ac.il/~snoop/Q.gif
Q is at one point
R is the radius of this cylinder, it's height is 2h
the cylinder is without the bases.
how can i calculate the electric flow through it?
the final answer is Q/[epsilon0*sqrt(1+R^2/h^2)]
where is my mistake:

we take ball with radius sqrt(R^2+h^2) and look on the rounded bases: the area of this ball inside the cylinder.

flow through bases / flow through all ball = bases area / all ball area

gaus: all ball flow is Q/epsilon0

all ball area is 4pi(R^2+h^2)

base area = circumference of projection of the base on y=2h * height of base
=(2pi*R)*[sqrt(R^2+h^2)-h]

2 bases area = base area * 2 = 4pi*R[sqrt(R^2+h^2)-h]

flow through bases=bases area*flow through all ball / all ball area=
= 4pi(R^2+h^2)Q/(epsilon0 4pi*R[sqrt(R^2+h^2)-h])=
Q(R^2+h^2)/(epsilon0 *R[sqrt(R^2+h^2)-h])

now that's not like that right answer, coz we can assign r=1 h=1
my answer qives 2/(sqrt(2)-1) * Q/epsilon0 = 2(1+sqrt(2)) * Q/epsilon0
right answer gives 1/sqrt(2) * Q/epsilon0

ori
ori is offline
#4
Oct30-04, 03:39 PM
P: 30

electric flow


Quote Quote by Tide
Why don't you try calculating the flux through the flat top and bottom of the cylinder? Then you could subtract it from the total flux given by Gauss' Law.
we get too hard integral at that case:
S r^2/(r^2+h^2)^(3/2) dr
or something like that


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