Please help homework question

please provide a solution

Note by Ishan Dixit
2 years, 6 months ago

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Using gauss theorem you can easily find the dependence of field on rr the distance from the axis.For this use E2πrh=ρπr2hE2πrh=\rhoπr^2h.The net flux through the part(cylinder) is Q/ϵQ/\epsilon.Now this is equal to ϕ1\phi1+ϕ2\phi2.To find ϕ2\phi2 or the flux through the circular face use the very definition of flux ϕ\phi=EdScosαEdScos\alpha.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/r21/r^2.So use the trivial method.

Spandan Senapati - 2 years, 6 months ago

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Answer is px(πa2)/2px(\pi*a^2)/2\in

Ishan Dixit - 2 years, 6 months ago

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but it isnt coming from this method please provide solution

Ishan Dixit - 2 years, 6 months ago

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@Ishan Dixit I have provided.If it weren't small then use integration.

Spandan Senapati - 2 years, 6 months ago

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@Spandan Senapati Thanks got it

Ishan Dixit - 2 years, 6 months ago

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@Ishan Dixit One thing how does the term R gets sorted ?

Ishan Dixit - 2 years, 6 months ago

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@Ishan Dixit I have changed it,the variables.Now its fine I guess.

Spandan Senapati - 2 years, 6 months ago

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@Spandan Senapati But how would the height of small cylinder that is x will come in answer

Ishan Dixit - 2 years, 6 months ago

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@Ishan Dixit Note that its r=xr=x and The Radius is R=aR=a.Substitute the variables now.OK?

Spandan Senapati - 2 years, 6 months ago

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@Spandan Senapati K

Ishan Dixit - 2 years, 6 months ago

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Yes its correct.Another thing worth noting is small circular face So we can neglect the small angle α>0\alpha-->0.So ϕ2\phi2=EπR2EπR^2=ρr/2ϵπR2\rho r/2\epsilon*πR^2.And Subtract this from ϕ\phi=ρπR2r/ϵ\rho πR^2r/\epsilon.The answer comes to be 1/21/2ϕ\phi

Spandan Senapati - 2 years, 6 months ago

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