Energy Stored in an Electric Field — Question 9

Question 9

A voltage-driven spring plate loses stability before the gap closes

A movable capacitor plate of area A is attached to a spring of stiffness K. Its zero-voltage gap is d0 and its displacement toward the fixed plate is x. At fixed V, take the attractive force as $F_e=\epsilon_0AV^2/[2(d_0-x)^2]$, for $0<x<d_0$.

1. Find the equilibrium relation between V and x.

2. Use force-slope stability to find the critical displacement and pull-in voltage.

Question 9 - Solution

A voltage-driven spring plate loses stability before the gap closes

(a) Method and calculation

$$Kx=\frac{\epsilon_0AV^2}{2(d_0-x)^2}.$$$$V^2=\frac{2K}{\epsilon_0A}x(d_0-x)^2.$$

(b) Interpretation and further calculation

$$\text{stable: }K>\frac{dF_e}{dx}=\frac{\epsilon_0AV^2}{(d_0-x)^3}=\frac{2Kx}{d_0-x}\Rightarrow x<d_0/3.$$$$x_{\mathrm{crit}}=d_0/3,\quad V_{\mathrm{pull}}=\sqrt{\frac{8Kd_0^3}{27\epsilon_0A}}.$$

Physical check

The unstable branch is not a second stable operating state; the ideal fixed-voltage force grows faster than the restoring force beyond the threshold.