Discussion
Halving the spacing of that capacitive transducer changes its capacitance to
Answer: A
Correct answer: 1.77 pF
Parallel plate capacitance is inversely proportional to the spacing, \[ C = \frac{\epsilon A}{d}, \] so halving \( d \) doubles \( C \): \( 2 \times 0.885 = 1.77 \) pF. That inverse relation makes the variable-gap capacitor very sensitive but distinctly non-linear, which is why such transducers are used over small gaps or in a differential arrangement.
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