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 $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.For the cube of

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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 $\phi = \frac{q_{enclosed}}{\varepsilon_0}$.For the cube of side length $a$,the length of the diagonal is $L_{cube} = a\sqrt{3}$. The enclosed charge is $q_1 = \lambda \cdot a\sqrt{3}$. Thus,the flux is $\phi_1 = \frac{\lambda a\sqrt{3}}{\varepsilon_0}$.For the sphere of radius $a$ (diameter $2a$),the length of the diameter is $L_{sphere} = 2a$. The enclosed charge is $q_2 = \lambda \cdot 2a$. Thus,the flux is $\phi_2 = \frac{\lambda \cdot 2a}{\varepsilon_0}$.The ratio of the flux is $\frac{\phi_1}{\phi_2} = \frac{\lambda a\sqrt{3} / \varepsilon_0}{\lambda \cdot 2a / \varepsilon_0} = \frac{\sqrt{3}}{2}$.

$A$ 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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