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A small sphere of radius $r_1$ and charge $q_1$ is enclosed by a spherical shell of radius $r_2$ and charge $q_2$. Show that if $q_1$ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge $\mathrm{q}_2$ on the shell is.
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Potential of inner sphere due to its own charge $=$ $\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1}$
Potential of inner sphere due to its presence inside the shell $=\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{r_2}$
$\therefore \quad$ Total potential of the sphere $=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$
Potential of the shell $=\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{r_2}$
Potential difference between sphere and shell $=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)-\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{r_2}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1}$
It is independent of the charge $\mathrm{q}_2$ on the shell. Since $\mathrm{q}_1$ is positive, potential difference between sphere and shell is positive. Charge (if positive) will always flow from sphere to shell.
Potential of inner sphere due to its presence inside the shell $=\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{r_2}$
$\therefore \quad$ Total potential of the sphere $=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$
Potential of the shell $=\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{r_2}$
Potential difference between sphere and shell $=\frac{1}{4 \pi \varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)-\frac{1}{4 \pi \varepsilon_0} \frac{q_2}{r_2}=\frac{1}{4 \pi \varepsilon_0} \frac{q_1}{r_1}$
It is independent of the charge $\mathrm{q}_2$ on the shell. Since $\mathrm{q}_1$ is positive, potential difference between sphere and shell is positive. Charge (if positive) will always flow from sphere to shell.
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