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GATE 2024 IN – Question 15

Sensors and Industrial Instrumentation · Resistive capacitive and inductive sensors · 1 mark · Multiple choice

The parallel-plate capacitor shown has plate area $A$. Its dielectric-layer thicknesses are $d_1,d_2$ and relative permittivities $\epsilon_1=1$ (air), $\epsilon_2$ (solid), respectively. Fringing is negligible; $\epsilon_0$ is free-space permittivity. If $d_1$ decreases by $\delta d_1$, the resultant capacitance is

Diagram for GATE 2024 IN question 15
  1. $\dfrac{\epsilon_0 A}{d_1-\delta d_1+d_2/\epsilon_2}$
  2. $\dfrac{\epsilon_0 A}{d_2+d_1/\epsilon_2}$
  3. $\dfrac{\epsilon_0 A}{d_2-\delta d_2+d_1/\epsilon_2}$
  4. $\dfrac{\epsilon_0 A}{d_1+\delta d_1+d_2/\epsilon_2}$

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Correct answer: (A) $\dfrac{\epsilon_0 A}{d_1-\delta d_1+d_2/\epsilon_2}$

Explanation

The air and solid dielectric layers act as series capacitors.

Thus $1/C=d_1/(\epsilon_0A)+d_2/(\epsilon_0\epsilon_2A)$. Replace $d_1$ by $d_1-\delta d_1$ to obtain $C=\epsilon_0A/(d_1-\delta d_1+d_2/\epsilon_2)$.

Answer: **A**.