# Electric flow

by ori
Tags: electric, flow
 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)]
 Sci Advisor HW Helper P: 3,144 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.
P: 30
 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

P: 30
Electric flow

 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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