Two point charges 8 mu text{C} and -2 mu text | vedclass

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

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$4 : 3$

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(C) According to Gauss's Law, the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.For the first sphere with radius $r_1 = 3 	ext{ cm}$, only the charge at $x = 2 	ext{ cm}$ $(q_1 = 8 \mu 	ext{C})$ is enclosed within the sphere.Therefore, the flux through the first sphere is $\phi_1 = \frac{8 \mu 	ext{C}}{\epsilon_0}$.For the second sphere with radius $r_2 = 5 	ext{ cm}$, both charges at $x = 2 	ext{ cm}$ $(8 \mu 	ext{C})$ and $x = 4 	ext{ cm}$ $(-2 \mu 	ext{C})$ are enclosed within the sphere.Therefore, the total enclosed charge is $q_{	ext{total}} = 8 \mu 	ext{C} - 2 \mu 	ext{C} = 6 \mu 	ext{C}$.The flux through the second sphere is $\phi_2 = \frac{6 \mu 	ext{C}}{\epsilon_0}$.The ratio of the electric flux is $\frac{\phi_1}{\phi_2} = \frac{8 \mu 	ext{C} / \epsilon_0}{6 \mu 	ext{C} / \epsilon_0} = \frac{8}{6} = \frac{4}{3}$.

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