Capacitor with a spring-mounted plate (electrostatic actuator)

One plate of a parallel-plate capacitor is a mass on a spring. The plate is held in place while a charge $Q$ is put on the capacitor and then released. It is the electrostatic counterpart of the electromagnet in problem 17, and the working principle of many MEMS sensors and actuators. The paper derives the equation of motion correctly and even solves it analytically. However, its parameters correspond to a gap of about 40 m and voltages above 100 MV, and its "small charge" plot shows numerical noise, not motion. The interactive simulator compares the isolated capacitor (constant charge) with a capacitor connected to a voltage source, which shows the snap-through ("pull-in") absent from the paper's case.

Problem statement

When there is no charge on the plates, the mass $m$ is at rest. In that position the mass is locked by a stop (Figure 1), and a charge $Q$ is put on the capacitor plates. When the stop is removed, the system starts to move. Determine the equation of motion of this electromechanical system.

A fixed plate and a movable plate (mass $m$) form a parallel-plate capacitor with area $S$ and initial gap $d$ (vacuum, $\varepsilon_0$).
The movable plate rolls without friction and is held by a spring $k$, unstretched at the start.
The displacement $x$ of the plate towards the fixed plate is the generalized coordinate. The gap is $d - x$.
After charging, the capacitor is disconnected, so the charge $Q$ stays constant.

Figure 1. Fixed plate (left) and movable plate of mass m on a spring k; gap d − x, area S, constant charge ±Q

Mathematical model

With the capacitor disconnected, the charge cannot change. The paper therefore uses the displacement $x$ as the only generalized coordinate, which is correct. The energies are:

paper$$\mathcal{R}^* = \tfrac12m\dot x^2,\qquad \mathcal{A} = \tfrac12kx^2,\qquad C(x) = \frac{\varepsilon_0S}{d-x},\qquad W_e = \frac{Q^2}{2C(x)} = \frac{Q^2(d-x)}{2\varepsilon_0S}$$

There is no inductor ($W_m^* = 0$) and no dissipation. The Lagrangian and the equation of motion are:

paper$$\mathcal{L} = \tfrac12m\dot x^2 - \tfrac12kx^2 - \frac{Q^2(d-x)}{2\varepsilon_0S}\qquad\Rightarrow\qquad m\ddot x + kx = \frac{Q^2}{2\varepsilon_0S} \equiv b$$

This is correct. At constant charge the field between the plates, $E = Q/(\varepsilon_0S)$, does not depend on the gap, so the attractive force $b = Q^2/(2\varepsilon_0S)$ is constant. The equation is linear, even though the capacitance changes with $x$.

Analytic solution

The paper solves the equation by the Laplace transform, with $x(0) = \dot x(0) = 0$:

$$X(s) = \frac{b}{m}\,\frac{1}{s\,(s^2 + k/m)}\qquad\Rightarrow\qquad x(t) = \frac bk\Big(1 - \cos\sqrt{\tfrac km}\,t\Big)$$

This is also correct. A constant force suddenly applied to an undamped spring–mass system makes it oscillate between $0$ and $2b/k$ around the static deflection $b/k$, with $\omega_n = \sqrt{k/m}$. Two consequences follow directly:

$x$ is never negative, whatever $Q$, $k$, $m$ or $S$.
Without damping the oscillation never settles.

The plates touch if $2b/k \ge d$, and the model then no longer applies.

Simulink model and results

The model integrates $\ddot x = -\frac kmx + \frac{q^2}{2\varepsilon_0Sm}$ with two integrators and a gain. An extension computes $C = \varepsilon_0S/(d-x)$. The paper verifies the simulation against the analytic solution, which is good practice.

Case	Parameters (ε₀ = 8.854·10⁻¹² F/m)	Paper's plot	Analytic
1	q = 10⁻⁵ C, k = 1, m = 0.5, S = 0.4, d ≈ 40 m	x between 0 and 28 m, C between 0.9 and 3·10⁻¹³ F	2b/k = 28.2 m, T = 4.44 s
2	q = 10⁻⁸ C, k = 0.1, d ≈ 1 m	x between 0 and 2.8·10⁻⁴ m	2b/k = 2.82·10⁻⁴ m, T = 14.0 s
3	q = 10⁻¹⁸ C, k = 10, 10 000 s	x oscillates about zero, ±1.7·10⁻⁴ m, amplitude growing in steps	2b/k = 2.8·10⁻²⁶ m

The gap $d$ is not stated. It follows from the capacitance plots: $C(0) = \varepsilon_0S/d$ gives $d \approx 40$ m in case 1 and $d\approx1$ m in case 2. The paper concludes:

A large charge prevents the plate from moving to negative elongation, while a small charge allows oscillation about zero (case 3).
Small charges need a very long settling time.
Larger $k$ raises the frequency, and $m$ acts the opposite way.
A larger $S$ gives a larger capacitance and a smaller amplitude.

Critical review

Case 3 shows numerical noise, not motion

With $q = 10^{-18}$ C the force is $b = 1.4\cdot10^{-25}$ N and the true motion stays below $2.8\cdot10^{-26}$ m, smaller than an atomic nucleus by many orders of magnitude. The plotted ±0.17 mm "oscillation" is integration error. Simulink's variable-step solver has a default absolute tolerance of $10^{-6}$, far above the signal. Each step adds an error of that order, and an undamped system keeps every error it receives. Over 10 000 s (about 7000 periods) the errors accumulate into the stepwise growing envelope seen in the plot. All conclusions drawn from this plot, about negative elongation, oscillation about zero and long settling, are therefore wrong. With absolute and relative tolerances appropriate to the signal (or scaled variables), the solution is $x = (b/k)(1-\cos\omega_nt)$ as in every other case.

"Large charge prevents negative elongation"

The analytic solution, derived in the paper itself, shows $x \ge 0$ for every charge. Starting from rest with a constant attractive force, the plate can never move away from the fixed plate past its starting point. Only an initial velocity or displacement could change that.

Physically impossible parameters

In case 1 the initial gap is about 40 m for plates of 0.4 m², and the plate moves 28 m. The parallel-plate formula assumes a gap much smaller than the plate size, so it is not valid here. The voltage is $U = Q/C = 10^{-5}/(0.9\cdot10^{-13}) \approx 1.1\cdot10^{8}$ V, far beyond the breakdown of any gap. Realistic electrostatic actuators have gaps of micrometres to millimetres and voltages of volts to hundreds of volts (preset "Realistic").

Settling, friction and the role of the parameters

The model has no damping, so nothing ever settles, for any charge. The slow growth in case 3 is the numerical drift described above.

Spring and mass: $k$ and $m$ set the frequency $\sqrt{k/m}$, as the paper says. The amplitude $2b/k$ depends on $k$ but not on $m$.
Plate area: a larger $S$ at constant charge reduces the field and therefore the force $b\propto1/S$.

The conclusion that a small charge gives "relatively large displacements because friction is neglected" mixes cause and effect. The displacement is $Q^2/(\varepsilon_0Sk)$, with or without friction. Static friction would simply prevent motion below a threshold force.

Smaller points

The gap $d$ is never stated. The conclusion calls the energy "electromagnetic" where it is electrostatic, and the system "hybrid" where "electromechanical" is meant. Nothing is said about the plates touching ($2b/k \ge d$).

What is goodThe reasoning for dropping $q$ as a coordinate (isolated capacitor, constant charge) is correct and clearly explained. The Lagrangian is right, the Laplace solution is right, and checking the simulation against the analytic solution is exactly what one should do.

Extension: constant charge versus constant voltage

If the capacitor stays connected to a voltage source $U$, the charge changes as the gap closes, $Q = C(x)U$. The force then grows as the gap shrinks:

$$\text{constant charge: } F = \frac{Q^2}{2\varepsilon_0S}\ \ (\text{constant}),\qquad \text{constant voltage: } F = \frac{\varepsilon_0SU^2}{2(d-x)^2}$$

In the constant-voltage case the energy method needs care. With $U$ prescribed, one uses the electric coenergy $W_e^* = \tfrac12C(x)U^2$ (the source does work). The force is $+\partial W_e^*/\partial x$, and not $-\partial W_e/\partial x$ at constant $U$. The two expressions agree at any instant, but they lead to different dynamics. At constant voltage the system has the same snap-through as the electromagnet in problem 17:

$$\text{stable equilibrium only for } x < \frac d3,\qquad U_\text{pi} = \sqrt{\frac{8kd^3}{27\,\varepsilon_0S}}$$

At constant charge there is no pull-in at all. The force does not grow as the gap closes, so the only limit is that the plates touch when $2Q^2/(2\varepsilon_0Sk) \ge d$. Driving a MEMS actuator with a charge source instead of a voltage source is a known technique to extend its stable travel beyond $d/3$.

Preset "Realistic": $S = 0.01$ m², $d = 1$ mm, $k = 2$ N/m, $m = 10$ g, charged to 100 V. Isolated, the plate oscillates between 0 and 0.44 mm. Connected to the same 100 V, it exceeds $U_\text{pi} = 82$ V and snaps shut.

Interactive simulator

The simulator integrates the equation with 4th-order Runge–Kutta in scaled variables, so it stays accurate even for the paper's tiny charges. A stop at a residual gap of 2 % of $d$ represents the plates touching.

Cases

Electrical condition

Compare with the other condition

Parameters

Q [C]

S [m²]

d [m]

k [N/m]

m [kg]

B [N·s/m]

Duration [s]

Number fields accept exponents, e.g. 1e-8.

Displacement x(t)

Capacitance C(t)

Electrostatic force F(t)

Static characteristic

Spring force kx and electrostatic force over the gap, for both conditions.

Animation (selected condition)

Results table

Review questions

Why is the force between the plates constant when the charge is constant?

The field of a parallel-plate capacitor, $E = Q/(\varepsilon_0S)$, depends only on the surface charge density, not on the gap. The force on one plate is $\tfrac12QE = Q^2/(2\varepsilon_0S)$.

Why can $x$ never become negative in the paper's set-up?

The solution is $x = (b/k)(1-\cos\omega_nt)$ with $b>0$. A constant attractive force applied from rest makes the plate oscillate between its starting point and $2b/k$.

How can you tell that a simulated result is numerical noise?

Compare its size with the scale of the physics (here $b/k \approx 10^{-26}$ m) and with the solver tolerances. A signal that grows over time in an undamped, linear, constant-coefficient system, without any physical driving term to explain it, is a red flag. Tighter tolerances or scaled variables make it disappear.

What changes if the capacitor stays connected to the voltage source?

The force becomes $\varepsilon_0SU^2/(2(d-x)^2)$ and grows as the gap closes. The plate snaps shut above $U_\text{pi} = \sqrt{8kd^3/(27\varepsilon_0S)}$, and stable positions exist only up to $d/3$.

Which energy expression is used at constant voltage, and why?

The coenergy $W_e^* = \tfrac12CU^2$ with $U$ as the given variable. The voltage source supplies energy as the capacitance changes, and the coenergy accounts for this, giving $F = \partial W_e^*/\partial x$.