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_{\

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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 radius $a$, the length of the diameter is $L_{	ext{sphere}} = 2a$. The charge enclosed is $q_{	ext{sphere}} = \lambda \cdot 2a$. Thus, the flux through the sphere is $\phi_{	ext{sphere}} = \frac{2\lambda a}{\epsilon_0}$.The ratio of the flux coming out of the cube to that of the sphere is $\frac{\phi_{	ext{cube}}}{\phi_{	ext{sphere}}} = \frac{\lambda a\sqrt{3} / \epsilon_0}{2\lambda a / \epsilon_0} = \frac{\sqrt{3}}{2}$.

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

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