Of the charge $Q$ on a tiny sphere, a fraction $\alpha$ is to be transferred to a second, nearby sphere. The spheres can be treated as particles. (a) What value of $\alpha$ maximizes the magnitude $F$ of the electrostatic force between the two spheres? What are the (b) smaller and (c) larger values of $\alpha$ that put $F$ at half the maximum magnitude?
Answers:
(a) $\alpha=0.5$
(b) $\alpha=0.15$
(c) $\alpha=0.85$
My attempt:
(a)
$q=\alpha Q$
$Q-q=Q-\alpha Q=Q(1-\alpha)$
$F=\frac{1}{4\pi\epsilon_0}\frac{\alpha(1-\alpha)Q^2}{r^2}$
$\frac{\partial F}{\partial \alpha}\left[ \frac{1}{4\pi\epsilon_0}\frac{\alpha(1-\alpha)Q^2}{r^2}\right]=0$
$Q^2\frac{\partial F}{\partial \alpha}[\alpha(1-\alpha)]=0$
$Q^2[-\alpha+1-\alpha]=Q^2(1-2\alpha)=0$
Solving for $\alpha$:
$1-2\alpha=0$
$\alpha=0.5$
For (b) and (c) I tried to solve $\frac{\alpha(1-\alpha)}{2}=0$ as a quadratic equation but didn't get the expected results. How should we solve this?
Let ${1\over4\pi\epsilon_0} \frac{Q^2}{r^2}$ be $k$, since it is a constant in the question. So, your expression of force becomes: $$F = k\alpha(1-\alpha) = k(-\alpha^2+\alpha)$$ Finding maxima is no big deal, and it was done correctly by you (although examinations may require you to show that the function has a maxima, although it is apparent since the leading coefficient of $-\alpha^2 + \alpha$ is negative). Getting to your actual question: I don't know how you got ${\alpha(1-\alpha)\over2}=0$, but the question asks what are the values of $\alpha$ that put $F$ at half the maximum magnitude. So, let find the maximum value of $F$: $$F = k\alpha(1-\alpha)$$ $$F_{max} = k\times0.5(1-0.5)$$ $$F_{max} = 0.25k$$ The question asks which values of $\alpha$ give $F = F_{max}/2$. Solving for $\alpha$ $$k(-\alpha^2+\alpha) = 0.25k/2$$ This yields: $$8\alpha^2 - 8\alpha+1 = 0$$ Solving by the quadratic formula, we get: $$\alpha_2 = \frac{2-\sqrt2}{4} = 0.146446609 \approx \color{green}{0.15}$$ $$\alpha_2 = \frac{2+\sqrt2}{4} = 0.85355339 \approx \color{green}{0.85}$$