$A$ 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 $q_{2}$ on the shell is.

(A) The potential $V_{1}$ of the inner sphere of radius $r_{1}$ is given by $V_{1} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}} \right)$.
The potential $V_{2}$ of the outer shell of radius $r_{2}$ is given by $V_{2} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{r_{2}} + \frac{q_{2}}{r_{2}} \right)$.
The potential difference between the sphere and the shell is $V = V_{1} - V_{2} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}} - \frac{q_{1}}{r_{2}} - \frac{q_{2}}{r_{2}} \right)$.
Simplifying this,we get $V = \frac{q_{1}}{4\pi\epsilon_{0}} \left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right)$.
Since $r_{2} > r_{1}$,the term $\left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right)$ is always positive.
If $q_{1} > 0$,then $V > 0$,which means $V_{1} > V_{2}$.
Therefore,charge will always flow from the inner sphere to the outer shell when connected by a wire,regardless of the value of $q_{2}$.