The linear charge density of a wire is 8.85,m | vedclass

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(C) According to Gauss's Law,the electric flux $\phi$ is given by $\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.Here,the wire passes through the cyl

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$4 	imes 10^6\, V\cdot m$

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(C) According to Gauss's Law,the electric flux $\phi$ is given by $\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.Here,the wire passes through the cylinder. The length of the wire enclosed within the cylinder is equal to the height of the cylinder,which is $L = 4\,m$.The enclosed charge is $q_{	ext{enclosed}} = \lambda 	imes L = 8.85 	imes 10^{-6}\,C/m 	imes 4\,m = 35.4 	imes 10^{-6}\,C$.The permittivity of free space is $\epsilon_0 \approx 8.85 	imes 10^{-12}\,C^2/(N\cdot m^2)$.Therefore,$\phi = \frac{35.4 	imes 10^{-6}}{8.85 	imes 10^{-12}} = 4 	imes 10^6\,V\cdot m$.

The linear charge density of a wire is $8.85\,\mu C/m$. The radius and height of the cylinder are $3\,m$ and $4\,m$ respectively. Find the electric flux passing through the cylinder.

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