The linear charge density of a wire is 8.85 | vedclass

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(D) According to the figure,the wire passes from the center of the top circular face to a point on the circumference of the bottom circular face. The

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$5 	imes 10^6 \ V-m$

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(D) According to the figure,the wire passes from the center of the top circular face to a point on the circumference of the bottom circular face. The length of the wire segment inside the cylinder $(L)$ forms the hypotenuse of a right-angled triangle with height $h = 4 \ m$ and base $r = 3 \ m$.$L = \sqrt{h^2 + r^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \ m$.The charge enclosed by the cylinder is $q_{	ext{enclosed}} = \lambda L = (8.85 	imes 10^{-6} \ C/m) 	imes (5 \ m) = 44.25 	imes 10^{-6} \ C$.According to Gauss's Law,the electric flux $\phi$ is given by $\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.Given $\epsilon_0 \approx 8.85 	imes 10^{-12} \ C^2/(N-m^2)$.$\phi = \frac{44.25 	imes 10^{-6}}{8.85 	imes 10^{-12}} = 5 	imes 10^6 \ V-m$.

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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