GATE 2025 IN Q.57: solution and answer

GATE 2025 · INQ.57 · 2 marks · NAT

An air filled parallel plate electrostatic actuator is shown in the figure. The area of each capacitor plate is $100 μ m \times 100 μ m$ . The distance between the plates $d_{0} = 1 μ m$ 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 ______ $\times 1 0^{- 14} C$ (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 $ε_{0} = 8.85 \times 1 0^{- 12} F/m$ and relative permittivity of air $(ε_{r})$ is 1.

GATE 2025 IN question 57: Two side-by-side sketches. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

The minimum charge is $Q = 2 ε_{0} A k d_{0} \approx 4.21 \times 1 0^{- 14} C$ , so the blank is 4.21.

Official answer: 4.00 to 4.40Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Plate area and the product we need: $A = (100 \times 1 0^{- 6} m)^{2} = 1 0^{- 8} m^{2}$ . With $ε_{0} = 8.85 \times 1 0^{- 12} F/m$ and $ε_{r} = 1$ , $ε_{0} A = 8.85 \times 1 0^{- 12} \times 1 0^{- 8} = 8.85 \times 1 0^{- 20} F \cdot m$ .
A current source sets the charge $Q$ on the plates, not the voltage, so this is charge control. At fixed $Q$ the stored energy is $W = \frac{Q ^{2}}{2 C} = \frac{Q ^{2} x}{2 ε _{0} A}$ , using $C = \frac{ε _{0} A}{x}$ with no fringing, where $x$ is the gap.
Force from virtual work at constant charge: $F_{e} = - \frac{\partial W}{\partial x}_{Q} = - \frac{Q ^{2}}{2 ε _{0} A}$ . The minus sign means it acts to shrink the gap (attraction), and its size does not depend on the gap. Check: the top plate sits in the field of the bottom plate alone, $E = \frac{σ}{2 ε _{0}}$ with $σ = \frac{Q}{A}$ , so $F = Q \cdot \frac{σ}{2 ε _{0}} = \frac{Q ^{2}}{2 ε _{0} A}$ .
Spring: once the top plate has moved down by $y$ , the spring pulls it back with $F_{s} = k y$ (Hooke's law), and this grows as the gap closes. The current source adds charge gradually, so the plate moves slowly through equilibrium positions with $\frac{Q ^{2}}{2 ε _{0} A} = k y$ at every instant. Because $F_{e}$ does not depend on the gap, there is no pull-in instability: the plate reaches the bottom plate ( $y = d_{0}$ ) only when the electrostatic force equals the largest spring force, $k d_{0}$ .
Condition for the gap to just close: $\frac{Q ^{2}}{2 ε _{0} A} = k d_{0} \Rightarrow Q = 2 ε_{0} A k d_{0}$ .
Substitute $k = 0.01 N/m$ and $d_{0} = 1 0^{- 6} m$ (so $k d_{0} = 1 0^{- 8} N$ ): $Q = 2 \times 8.85 \times 1 0^{- 20} \times 0.01 \times 1 0^{- 6} = 1.77 \times 1 0^{- 27} = 4.207 \times 1 0^{- 14} C$
To two decimal places, $Q \approx 4.21 \times 1 0^{- 14} C$ , so the blank is $4.21$ .

KEY IDEA

A current source fixes the charge, not the voltage. With charge fixed, the pull between the plates is $\frac{Q ^{2}}{2 ε _{0} A}$ whatever the gap. The spring's pull-back $k y$ grows as the gap closes and is largest at contact, $k d_{0}$ . So the gap closes when $\frac{Q ^{2}}{2 ε _{0} A} = k d_{0}$ .

WATCH OUT FOR

Using the voltage-drive result. At fixed voltage the force $\frac{ε _{0} A V ^{2}}{2 x ^{2}}$ grows as the gap shrinks, which gives the familiar pull-in at one third of the gap. Setting the balance at $y = d_{0} /3$ gives $Q \approx 2.43 \times 1 0^{- 14} C$ , which is wrong here: at fixed charge the force is constant and there is no pull-in. A second trap is treating the charge as switched on suddenly. A constant force applied in one step makes the plate overshoot to twice its equilibrium displacement, which gives $Q = ε_{0} A k d_{0} \approx 2.97 \times 1 0^{- 14} C$ . A current source builds the charge up gradually, so there is no overshoot.

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GATE 2025 IN Q.57: solution and answer

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