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A positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is flux through closed gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is flux through closed gaussian surface $S_2$ whose radius is just greater than $R_1$ then:-
$\phi_1 > \phi_2$
$\phi_2 > \phi_1$
$\phi_1 = \phi_2 = \frac {q}{\varepsilon_0}$
$\phi_1 = \phi_2 = \frac {kq}{\varepsilon_0}$
Solution
Flux through
$\mathrm{S}_{1}\left(\phi_{1}\right)=\frac{\text { Charge enclosed within } \mathrm{S}_{1}}{\varepsilon_{o}}=\frac{q}{\varepsilon_{o}}$
Since the shell is conducting therefore a charge of magnitude $-$ $q$ will be induced at the inner surface (radius $\mathrm{R}_{1}$ ) of the shell.
$\therefore$ Flux through
$\mathrm{S}_{2}\left(\phi_{2}\right)=\frac{\text { Charge enclosed within } \mathrm{S}_{2}}{\varepsilon_{\mathrm{o}}}=\frac{q-q}{\varepsilon_{o}}=0$
