GnitGnit
ContestsProblemsBlogSubmit a ProblemMy PageSign In

© 2026 Gnit. All rights reserved.

Terms of ServicePrivacy PolicyThird-Party SoftwareContactOfficial X
Contests/Gnit Sunday Open 002 (GSO002)/Problem 7 Pull-In Effect Analysis in a MEMS Actuator
Problem 7

Pull-In Effect Analysis in a MEMS Actuator

Finished
200 ptsLv.4 ElementaryElectromagnetism
2026/03/22 19:30〜2026/03/22 21:00
Author: admin

Problem Statement

In microelectromechanical systems (MEMS), electrostatic actuators that use electrostatic force to control movable parts are widely used. Consider the physical instability (pull-in effect) inherent in such a system.

A fixed lower plate of area SSS is placed horizontally. Directly above it, an upper plate (movable plate) of area SSS and mass mmm is suspended by a light insulating spring with spring constant kkk. The two plates form a parallel plate capacitor facing each other. The space between the plates is vacuum with permittivity ε0\varepsilon_0ε0​. Edge effects at the plate boundaries can be neglected.

Initially, no voltage is applied between the plates, and the movable plate is at rest in equilibrium between gravity and the spring's elastic force. At this point, the plate separation is ddd and the capacitance of the capacitor is C0C_0C0​.

From this state, a DC power supply is connected between the plates, and the applied voltage VVV is slowly increased from 000 (slowly enough that the plates do not vibrate, always maintaining force equilibrium). As the voltage increases, the plate separation decreases. When the voltage VVV exceeds a critical value V0V_0V0​, the spring's restoring force can no longer support the electrostatic force, and the movable plate is suddenly attracted to and collides with the lower plate. This phenomenon is called the "pull-in effect," and the critical voltage V0V_0V0​ is called the "pull-in voltage."

Find the value of V02V_0^2V02​, the square of the pull-in voltage.

Constraints

  • Spring constant: k=1.35 N/mk = 1.35 \text{ N/m}k=1.35 N/m
  • Initial plate separation: d=6.4×10−6 md = 6.4 \times 10^{-6} \text{ m}d=6.4×10−6 m
  • Initial capacitance: C0=2.4×10−12 FC_0 = 2.4 \times 10^{-12} \text{ F}C0​=2.4×10−12 F
  • Mass: m=2.0×10−6 kgm = 2.0 \times 10^{-6} \text{ kg}m=2.0×10−6 kg
  • Gravitational acceleration: g=9.8 m/s2g = 9.8 \text{ m/s}^2g=9.8 m/s2
  • All physical quantities are in SI units under ideal vacuum conditions.

Input Format

The square of the pull-in voltage V02V_0^2V02​ (in units of V2\text{V}^2V2) is expressed as an irreducible fraction AB\frac{A}{B}BA​. Give the value of A+BA + BA+B as a positive integer.

Submit Answer

Please sign in to submit an answer

Sign In

Calculator

0
View Scoreboard
1234567891011121314151617181920
Read Solution Blog