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(A) According to Gauss's Law,the electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.For th

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$\phi_1 > \phi_2$

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(A) According to Gauss's Law,the electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.For the Gaussian surface $S_1$,the charge enclosed is simply the point charge $q$. Thus,$\phi_1 = \frac{q}{\varepsilon_0}$.For the Gaussian surface $S_2$,which lies within the conducting material of the shell (just greater than $R_1$),the surface encloses both the point charge $q$ at the center and the induced charge $-q$ on the inner surface of the conducting shell. The total charge enclosed is $q_{	ext{enclosed}} = q + (-q) = 0$.Therefore,$\phi_2 = \frac{0}{\varepsilon_0} = 0$.Comparing the two,we find that $\phi_1 = \frac{q}{\varepsilon_0}$ and $\phi_2 = 0$,which implies $\phi_1 > \phi_2$.

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