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.
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.
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Correct Answer: 4
Solution and Explanation
The electrostatic force between the capacitor plates must equal the spring restoring force:
\[ F_{\text{electrostatic}} = F_{\text{spring}} \]
Step 2: Expressions for forces.
Electrostatic force:
\[ F = \frac{Q^2}{2 \varepsilon A} \]
Spring force:
\[ F = k d_0 \]
Step 3: Equating forces.
\[ \frac{Q^2}{2 \varepsilon A} = k d_0 \]
\[ \Rightarrow Q^2 = 2 \varepsilon A k d_0 \]
Step 4: Substituting values.
Area:
\[ A = (100 \times 10^{-6})^2 = 10^{-8}~\text{m}^2 \]
Permittivity:
\[ \varepsilon = \varepsilon_0 = 8.85 \times 10^{-12}~\text{F/m} \]
Spring constant and initial displacement:
\[ k = 0.01~\text{N/m}, \quad d_0 = 1 \times 10^{-6}~\text{m} \]
Substitute into the equation:
\[ Q^2 = 2 \times 8.85 \times 10^{-12} \times 10^{-8} \times 0.01 \times 10^{-6} = 1.77 \times 10^{-27} \]
Taking square root:
\[ Q = \sqrt{1.77 \times 10^{-27}} \approx 4.2 \times 10^{-14}~\text{C} \]
Final Answer:
Q = 4 × 10−14 C
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