Question

An air filled parallel plate electrostatic actuator is shown in the figure. The area of each capacitor plate is $100 \mu m \times 100 \mu m$. The distance between the plates $d_0 = 1 \mu m$ when both the capacitor charge and spring restoring force are zero as shown in Figure (a). A linear spring of constant $k = 0.01 N/m$ is connected to the movable plate. When charge is supplied to the capacitor using a current source, the top plate moves as shown in Figure (b). The magnitude of minimum charge (Q) required to momentarily close the gap between the plates is ________ $\times 10^{-14}$ C (rounded off to two decimal places).
Note: Assume a full range of motion is possible for the top plate and there is no fringe capacitance. The permittivity of free space is $\epsilon_0 = 8.85 \times 10^{-12}$ F/m and relative permittivity of air ($\epsilon_r$) is 1.

Hint:

The minimum charge required to initiate the closure of the gap in an electrostatic actuator with a spring is determined by the pull-in voltage or charge, which occurs at a specific fraction of the initial gap.

Answer 1

Step 1: Force balance at equilibrium.
The electrostatic force between the capacitor plates must equal the spring restoring force:
\[ F_{{electrostatic}} = F_{{spring}} \] Step 2: Expressions for forces.
Electrostatic force:
\[ F = \frac{Q^2}{2 \varepsilon A} \] Spring force:
\[ F = k d_0 \] Step 3: Equating forces:
\[ \frac{Q^2}{2 \varepsilon A} = k d_0 \quad \Rightarrow \quad Q^2 = 2 \varepsilon A k d_0 \] Step 4: Substituting values:

 \[ Q^2 = 2 \cdot 8.85 \times 10^{-12} \cdot 10^{-8} \cdot 0.01 \cdot 10^{-6} = 1.77 \times 10^{-27} \] \[ Q = \sqrt{1.77 \times 10^{-27}} \approx 4.2 \times 10^{-14} \, {C} \] Final Answer:
\[ \boxed{Q = 4 \times 10^{-14} \, {C}} \]