Question

A parallel plate capacitive displacement sensor has a plate area of \(2\ \text{cm}^2\). The air gap between the plates is decreased by \(0.1\) mm from an initial value of \(0.5\) mm. The percentage change in the capacitance value is __________ %.
(Assume permittivity as \(\epsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\))

• A. 10.0
• B. 14.1
• C. 25.0
• D. 30.0

Hint:

Capacitance of a parallel plate sensor is \(C = \epsilon_0 A/d\), so the percentage change only depends on the ratio of the old gap to the new gap, the area and permittivity cancel out.

Answer 1

The Correct Option is C

Step 1: Write the capacitance formula for a parallel plate sensor.
For a parallel plate capacitor with plate area \(A\), air gap \(d\), and permittivity of free space \(\epsilon_0\) (air-filled gap, relative permittivity 1), the capacitance is
\[ C = \frac{\epsilon_0 A}{d} \]
Capacitance is inversely proportional to the gap \(d\), a smaller gap gives a larger capacitance, this is exactly how a capacitive displacement sensor turns a change in gap into a change in an electrical quantity.

Step 2: Identify the initial and final gap.
The initial gap is \(d_1 = 0.5\) mm. The gap decreases by \(0.1\) mm, so the new gap is
\[ d_2 = 0.5 - 0.1 = 0.4\ \text{mm} \]

Step 3: Write the ratio of the two capacitances.
Since \(A\) and \(\epsilon_0\) do not change, they cancel out of the ratio:
\[ \frac{C_2}{C_1} = \frac{\epsilon_0 A/d_2}{\epsilon_0 A/d_1} = \frac{d_1}{d_2} \]
This is why the exact value of \(\epsilon_0 = 8.854\times10^{-12}\) F/m and the plate area \(2\ \text{cm}^2\) given in the question do not actually need to be substituted, a PERCENTAGE change only depends on the ratio of the gaps.

Step 4: Compute the ratio and the percentage change.
\[ \frac{C_2}{C_1} = \frac{0.5}{0.4} = 1.25 \]
The percentage change in capacitance is
\[ \% \text{change} = \left(\frac{C_2-C_1}{C_1}\right)\times 100 = \left(\frac{C_2}{C_1}-1\right)\times 100 = (1.25-1)\times 100 = 25.0\% \]

Step 5: Why the other options are wrong.
10.0% (option A) would come from wrongly computing the fractional change in the gap itself, \(0.1/1.0\), instead of the resulting change in capacitance. 14.1% (option B) is the kind of number that appears when a \(\sqrt2\) factor is mixed in from an unrelated AC-bridge or RMS-type calculation, it has no place in this simple inverse relationship. 30.0% (option D) would follow from an algebra slip in the ratio and does not match the exact calculation above.

Final Answer:
The capacitance increases by 25.0% when the gap decreases from 0.5 mm to 0.4 mm. \[ \boxed{\%\text{change} = 25.0\%} \]