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Question Number 34205 by rahul 19 last updated on 02/May/18

Commented by rahul 19 last updated on 02/May/18

$$ans.givenis\frac{\rho x}{2{ϵ}_0}\left(\pi a^2\right).$$

Answered by ajfour last updated on 03/May/18

$$totalflux{\varphi }_{tot}=\rho \left(\pi a^{2}x\right)/{ϵ}_0$$$$fieldatr=xbeE,then$$$${ϵ}_{0}E\left(2\pi rl\right)=\rho \left(\pi r^{2}l\right)$$$$\Rightarrow E=\frac{\rho r}{2{ϵ}_0}$$$$fluxthroughflatcirculararea$$$$ofsmallcylinderatr=xis$$$${\varphi }_{flatcircular}=E\left(\pi a^2\right)=\frac{\rho x}{2{ϵ}_0}\left(\pi a^2\right)$$$${\varphi }_{curvedsurface}={\varphi }_{tot}-{\varphi }_{flatcircular}$$$$=\frac{\rho x}{{ϵ}_0}\left(\pi a^2\right)-\frac{\rho x}{2{ϵ}_0}\left(\pi a^2\right)$$$$\Rightarrow {\mathit{\varphi }}_{curved}=\frac{\rho x}{2{ϵ}_0}\left(\pi a^2\right).$$

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