= Solution
The <image charge> construction replaces conductor-induced charge by fictitious sources outside the physical domain. Match the conductor potentials and the far-field condition; uniqueness then fixes the potential in the physical region. For a grounded plane, reflect a charge and reverse its sign, giving attraction $q^2/(16\pi\epsilon_0h^2)$. For a grounded sphere of radius $a$, a charge at distance $R>a$ has image $-qa/R$ at radial distance $a^2/R$. These are examples with prescribed zero potential, not arbitrary isolated conductors.<Method of Images by Richard Fitzpatrick>[grounded plane construction and its uniqueness argument]{ref}
Use horizontal $x$ and upward $y$ in the upper right-angle wedge. The sheets meet electrically, so take their common potential as zero and assume no externally imposed field. For a real charge at $(0,y)$, the <image charges in a right-angle conducting wedge> are $-q$ at $(y,0)$ and $(-y,0)$, and $+q$ at $(0,-y)$. The first two contributions have cancelling horizontal components. With $K_e=1/(4\pi\epsilon_0)$,
$$
F_y=-\frac{K_eq^2}{\sqrt2y^2}+\frac{K_eq^2}{4y^2}
=-\frac\beta{y^2},\qquad
\boxed{\beta=\frac{q^2(2\sqrt2-1)}{16\pi\epsilon_0}.}
$$
Thus the <force> magnitude at the original height is $\beta/y_0^2$, directed down toward the intersection. The sphere is treated as a <point charge>, with size much smaller than its distances from the sheets.
Since gravity is already balanced in the uncharged <equilibrium>, the charged vertical <equilibrium> satisfies $k(y_e-y_0)+\beta/y_e^2=0$. For $\beta/(ky_0^3)\ll1$, $y_e\simeq y_0-\beta/(ky_0^2)$. Differentiating the <force> gives
$$
\omega_y^2=\frac1m\left(k-\frac{2\beta}{y_e^3}\right),\qquad
\boxed{\frac{\omega_y-\omega_0}{\omega_0}\simeq-\frac{\beta}{ky_0^3},\quad\omega_0=\sqrt{k/m}.}
$$
The <electrostatic softening above a conducting wedge> reduces the restoring <force>; the small-charge vertical <equilibrium> remains stable.
For the <lateral stability of a spring-supported charge in a wedge>, the off-axis images of $(x,y)$ are $-q$ at $(y,x)$ and $(-y,-x)$ and $+q$ at $(-x,-y)$. Expansion near the bisector gives
$$
F_x=D_xx+O(x^3),\qquad
D_x=\frac{K_eq^2}{y_e^3}\left(\sqrt2+\frac14\right)>0.
$$
So \b[the electrostatic <force> is laterally destabilizing]: moving toward either sheet strengthens its attraction. In the simplified model where the <spring> supplies only a vertical <force>, the overall <equilibrium> is a saddle and is unstable.
A three-dimensional <spring> suspension also has geometrical transverse stiffness. If its charged-equilibrium tension is $T$ and <length> is $\ell$, a small sideways displacement gives <spring> <force> $-(T/\ell)x$, with $T=mg+\beta/y_e^2$. The full stability conditions in this suspension model are
$$
\boxed{k>\frac{2\beta}{y_e^3},\qquad\frac T\ell>D_x.}
$$
Motion parallel to the intersection line has no electrostatic stiffness in the ideal infinite-wedge geometry and is stabilized by the positive <spring> tension. Thus \b[lateral electrostatic destabilization alone does not prove instability of a real hanging spring]. At fixed finite support geometry, a sufficiently small charge preserves stability because $D_x=O(q^2)$ while $T/\ell\to mg/\ell>0$. The given vertical-displacement approximation does not specify the attachment <length> or by itself establish the transverse inequality for an arbitrary charge. The general criterion above states what can be concluded; the saddle verdict applies to the distinct idealization with zero lateral support stiffness.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1-image-wedge.png]
{title=Original image-charge field construction and the opposite signs of vertical softening and lateral electrostatic stiffness}
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