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electric flow |
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| Oct30-04, 03:19 PM | #1 |
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electric flow
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)] |
| Oct30-04, 03:30 PM | #2 |
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Recognitions:
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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.
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| Oct30-04, 03:38 PM | #3 |
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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 |
| Oct30-04, 03:39 PM | #4 |
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electric flowS r^2/(r^2+h^2)^(3/2) dr or something like that |
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