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.

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 \(1/2\)\(\phi\)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] 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

Sort by:

TopNewestUsing 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.

Log in to reply

Answer is \(px(\pi*a^2)/2\in\)

Log in to reply

but it isnt coming from this method please provide solution

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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 \(1/2\)\(\phi\)

Log in to reply