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 \(\epsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}\))
Show 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.
A different route: compute both capacitances in actual farads and compare them directly. Convert every quantity to base SI units.
Plate area: $A = 2\ \text{cm}^2 = 2\times10^{-4}\ \text{m}^2$. Initial gap: $d_1 = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}$. New gap: $d_2 = 0.5-0.1 = 0.4\ \text{mm} = 4\times10^{-4}\ \text{m}$. Permittivity: $\epsilon_0 = 8.854\times10^{-12}\ \text{F/m}$.
Compute the initial capacitance.
\[
C_1 = \frac{\epsilon_0 A}{d_1} = \frac{8.854\times10^{-12}\times 2\times10^{-4}}{5\times10^{-4}}
\]
\[
C_1 = \frac{1.7708\times10^{-15}}{5\times10^{-4}} = 3.5416\times10^{-12}\ \text{F} = 3.5416\ \text{pF}
\]
Compute the new capacitance.
\[
C_2 = \frac{\epsilon_0 A}{d_2} = \frac{8.854\times10^{-12}\times 2\times10^{-4}}{4\times10^{-4}}
\]
\[
C_2 = \frac{1.7708\times10^{-15}}{4\times10^{-4}} = 4.4270\times10^{-12}\ \text{F} = 4.4270\ \text{pF}
\]
Compute the percentage change from the actual numbers.
\[
\%\text{change} = \frac{C_2-C_1}{C_1}\times 100 = \frac{4.4270-3.5416}{3.5416}\times 100
\]
\[
= \frac{0.8854}{3.5416}\times 100 \approx 25.0\%
\]
Cross-check.
Working with the full numeric capacitances in picofarads, instead of the pure gap ratio, gives the same 25.0% answer, confirming that the area and permittivity values given in the question, while not strictly necessary once you notice they cancel, are consistent and do not change the result.
\[ \boxed{\%\text{change} = 25.0\%} \]
