Image charges in a right-angle conducting wedge ($F_y=-q^2(2\sqrt{2}-1)/(16\pi {ϵ}_0{y}^2)$)

For conducting sheets $y=\pm x$ at common zero potential and a charge in $y>|x|$, successive reflections give two negative images and one positive double-reflection image. On the bisector, the negative images lie at $(\pm y,0)$ and the positive image at $(0,-y)$. Summing their fields gives the displayed downward force. Image forces exclude the real charge's self-field; replacing the induced charge by images does not make the images physical particles.

The electrostatic bisector force is laterally anti-restoring, with the positive stiffness coefficient on the right. A spring of tension $T$ and length $\ell$ adds transverse restoring stiffness $T/\ell$ when its fixed support is directly above the charge. The displayed inequality therefore distinguishes actual transverse stability from the saddle of a model with only a vertical spring force. Axial spring stiffness alone does not specify the transverse verdict. At fixed finite support geometry and positive gravitational load, sufficiently small electric charge preserves lateral stability: the destabilizing coefficient tends to zero as $q^2$, while the tension term approaches $mg/\ell >0$. The vertical equilibrium must separately satisfy its own positive-stiffness condition.

For an attractive force $-\beta /y^2$ and an otherwise linear vertical spring, differentiation about the charged equilibrium $y_e$ reduces the spring stiffness by $2\beta /y_e^3$. If the uncharged height is $y_0$ and the shift is small, the fractional frequency change is $-\beta /(k{y}_0^3)$. Using only the force value, without its derivative, would miss this frequency shift.