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Please help homework question

please provide a solution

Note by Ishan Dixit
4 months ago

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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 \(E2πrh=\rhoπr^2h\).The net flux through the part(cylinder) is \(Q/\epsilon\).Now this is equal to \(\phi1\)+\(\phi2\).To find \(\phi2\) or the flux through the circular face use the very definition of flux \(\phi\)=\(EdScos\alpha\).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 \(1/r^2\).So use the trivial method. Spandan Senapati · 4 months ago

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@Spandan Senapati Answer is \(px(\pi*a^2)/2\in\) Ishan Dixit · 4 months ago

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@Ishan Dixit Yes its correct.Another thing worth noting is small circular face So we can neglect the small angle \(\alpha-->0\).So \(\phi2\)=\(EπR^2\)=\(\rho r/2\epsilon*πR^2\).And Subtract this from \(\phi\)=\(\rho πR^2r/\epsilon\).The answer comes to be \(1/2\)\(\phi\) Spandan Senapati · 4 months ago

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@Ishan Dixit but it isnt coming from this method please provide solution Ishan Dixit · 4 months ago

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@Ishan Dixit I have provided.If it weren't small then use integration. Spandan Senapati · 4 months ago

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@Spandan Senapati Thanks got it Ishan Dixit · 4 months ago

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@Ishan Dixit One thing how does the term R gets sorted ? Ishan Dixit · 4 months ago

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@Ishan Dixit I have changed it,the variables.Now its fine I guess. Spandan Senapati · 4 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 · 4 months ago

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@Ishan Dixit Note that its \(r=x\) and The Radius is \(R=a\).Substitute the variables now.OK? Spandan Senapati · 4 months ago

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@Spandan Senapati K Ishan Dixit · 4 months ago

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