= Solution
{c}
Put $C=q^2/(4\pi\epsilon_0)>0$. The <spring> exerts the <force> $k(\ell-x)$ by <Hooke's law>, while <Coulomb's law> gives the attractive <force> $-C/x^2$. <Newton's second law> therefore gives the <equation of motion>
$$
\boxed{m\ddot x=f(x)=k(\ell-x)-\frac C{x^2}.}
$$
For $x>0$, the <derivative> is $f'(x)=-k+2C/x^3$. Thus $f$ increases up to $x_*=(2C/k)^{1/3}$ and decreases thereafter. Also $f(x)\to-\infty$ at both ends of the positive half-axis. Its unique <maximum> is
$$
f(x_*)=k\left(\ell-\frac32x_*\right).
$$
There are therefore two <equilibrium points> precisely when
$$
\boxed{\ell>\ell_1=\frac32\left(\frac{q^2}{2\pi\epsilon_0k}\right)^{1/3}.}
$$
Writing them $x_-<x_*<x_+$, the <linearization> $m\ddot\eta=f'(x_\pm)\eta$ has exponentially growing solutions at $x_-$ and oscillatory solutions at $x_+$. Equivalently, the <potential energy> has a strict <local maximum> at $x_-$ and a strict <local minimum> at $x_+$. Hence \b[the smaller <equilibrium point> is an <unstable equilibrium>; the larger is a <stable equilibrium>], with no <damping> to make it attracting.
At equality $\ell=\ell_1$ there is one degenerate <equilibrium point> $x_*$; it is unstable, since a displacement to its left gives a leftward <force>. For $\ell<\ell_1$ there are no <equilibrium points>.
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