  Note by Ishan Dixit
4 years, 3 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.

- 4 years, 3 months ago

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

- 4 years, 3 months ago

but it isnt coming from this method please provide solution

- 4 years, 3 months ago

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

- 4 years, 3 months ago

Thanks got it

- 4 years, 3 months ago

One thing how does the term R gets sorted ?

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

Note that its $r=x$ and The Radius is $R=a$.Substitute the variables now.OK?

- 4 years, 3 months ago

K

- 4 years, 3 months ago

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$

- 4 years, 3 months ago