GATE 2025 IN – Question 57
The shown air-filled parallel-plate electrostatic actuator has plate area $100\ \mu\mathrm m\times100\ \mu\mathrm m$. With zero charge and spring restoring force, the plate gap is $d_0=1\ \mu\mathrm m$, as in Figure (a). A linear spring of stiffness $k=0.01$ N/m is attached to the movable top plate. A current source supplies charge and moves it as in Figure (b). The minimum charge magnitude to momentarily close the gap is ______ $\times10^{-14}$ C (rounded to two decimal places). Assume full plate motion and no fringe capacitance. Given: $\epsilon_0=8.85\times10^{-12}$ F/m and air relative permittivity 1.
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Show answer and explanation
Correct answer: 4.00 to 4.40
Explanation
For fixed charge, electrostatic attraction is $F=Q^2/(2\epsilon_0A)$, independent of the remaining gap.
Under the quasi-static closing interpretation used by the key, the threshold satisfies $F=kd_0$.
Thus $Q=\sqrt{2\epsilon_0Akd_0}=\sqrt{2(8.85\times10^{-12})(10^{-8})(0.01)(10^{-6})}=4.2071\times10^{-14}$ C.
The coefficient is **4.21**.
A dynamic charge-step interpretation would require the motion/charge history; the stated result uses the quasi-static threshold.