A linear charge having linear charge density | vedclass

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(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{inside}}}{\epsilon_0}$,where $q_{	e

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$\frac{\sqrt{3}}{2}$

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(C) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{inside}}}{\epsilon_0}$,where $q_{	ext{inside}}$ is the total charge enclosed by the surface.For the cube of side length $a$,the length of the diagonal is $L_{	ext{cube}} = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$. The charge enclosed is $q_{	ext{cube}} = \lambda \cdot a\sqrt{3}$. Thus,the flux through the cube is $\phi_{	ext{cube}} = \frac{\lambda a\sqrt{3}}{\epsilon_0}$.For the sphere of diameter $a$ (radius $R = a/2$),the length of the diameter is $L_{	ext{sphere}} = a$. The charge enclosed is $q_{	ext{sphere}} = \lambda \cdot a$. Thus,the flux through the sphere is $\phi_{	ext{sphere}} = \frac{\lambda a}{\epsilon_0}$.The ratio of the flux coming out of the cube to that of the sphere is $\frac{\phi_{	ext{cube}}}{\phi_{	ext{sphere}}} = \frac{\lambda a\sqrt{3} / \epsilon_0}{\lambda a / \epsilon_0} = \sqrt{3} : 1$ or $\frac{\sqrt{3}}{1}$.Wait,re-evaluating the provided options: if the sphere diameter is $a$,the ratio is $\sqrt{3}$. If the sphere radius is $a$,the diameter is $2a$,then the ratio is $\frac{\sqrt{3}a}{2a} = \frac{\sqrt{3}}{2}$. Given the options,it is assumed the sphere has a radius $a$ (diameter $2a$). Thus,$\phi_{	ext{sphere}} = \frac{\lambda (2a)}{\epsilon_0}$.Ratio = $\frac{\lambda a\sqrt{3}}{\lambda (2a)} = \frac{\sqrt{3}}{2}$.

$A$ linear charge having linear charge density $\lambda$ penetrates a cube diagonally and then it penetrates a sphere diametrically as shown. What will be the ratio of flux coming out of the cube and the sphere?

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