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2 \times 3
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a_{i-1}
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Using gauss theorem you can easily find the dependence of field on r the distance from the axis.For this use E2πrh=ρπr2h.The net flux through the part(cylinder) is Q/ϵ.Now this is equal to ϕ1+ϕ2.To find ϕ2 or the flux through the circular face use the very definition of flux ϕ=EdScosα.And note that the field that we computed is radially outwards.An important aspect to note is that one can't use Solid Angle here as the field dependence isn't 1/r2.So use the trivial method.
Yes its correct.Another thing worth noting is small circular face So we can neglect the small angle α−−>0.So ϕ2=EπR2=ρr/2ϵ∗πR2.And Subtract this from ϕ=ρπR2r/ϵ.The answer comes to be 1/2ϕ
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Using gauss theorem you can easily find the dependence of field on r the distance from the axis.For this use E2πrh=ρπr2h.The net flux through the part(cylinder) is Q/ϵ.Now this is equal to ϕ1+ϕ2.To find ϕ2 or the flux through the circular face use the very definition of flux ϕ=EdScosα.And note that the field that we computed is radially outwards.An important aspect to note is that one can't use Solid Angle here as the field dependence isn't 1/r2.So use the trivial method.
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Answer is px(π∗a2)/2∈
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but it isnt coming from this method please provide solution
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r=x and The Radius is R=a.Substitute the variables now.OK?
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Yes its correct.Another thing worth noting is small circular face So we can neglect the small angle α−−>0.So ϕ2=EπR2=ρr/2ϵ∗πR2.And Subtract this from ϕ=ρπR2r/ϵ.The answer comes to be 1/2ϕ