Answer
$\alpha=\frac{1}{2}$
Work Step by Step
Of the charge $Q $on a tiny sphere, a fraction $\alpha$ is to be transferred to a second, nearby sphere.
Therefore, charge of second sphere is: $\alpha Q$
and charge of second sphere is: $Q(1-\alpha)$
Let $r$ is the distance between the spheres.
The magnitude of electrostatic force between them is given by
$F=\frac{1}{4\pi\epsilon_0}\frac{\alpha Q^2(1-\alpha)}{r^2}$
Differentiating both side with respect to $\alpha$, we obtain
$\frac{dF}{d\alpha}=\frac{Q^2}{4\pi\epsilon_0r^2}(1-2\alpha)$
For maximum value of $F$,
$\frac{dF}{d\alpha}=0$
or, $\frac{Q^2}{4\pi\epsilon_0r^2}(1-2\alpha)=0$
or, $(1-2\alpha)=0$
or, $\alpha=\frac{1}{2}$
