Capacitive Displacement Transducer – Step by Step Solution

Given

• Plate area = \(50\,mm \times 50\,mm\)
• Plate spacing \(d = 0.5\,mm\)
• Change in capacitance \(\Delta C = 10\,pF\)
• Permittivity of air \[ \varepsilon_0 = 8.854 \times 10^{-12} \, F/m \]

Sensitivity definition:

\[ S=\frac{\Delta C}{\Delta x} \]

For a parallel plate capacitive displacement transducer:

\[ C=\frac{\varepsilon A}{d} \]

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Step 1: Convert Units

Area

\[ A=50\,mm \times 50\,mm \] \[ A=0.05\,m \times 0.05\,m \] \[ A=0.0025\,m^2 \]

Distance

\[ d=0.5\,mm \] \[ d=0.5\times10^{-3} \] \[ d=5\times10^{-4}\,m \]

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Step 2: Find Initial Capacitance

\[ C=\varepsilon_0\frac{A}{d} \]

Substitute values:

\[ C=8.854\times10^{-12}\times\frac{0.0025}{5\times10^{-4}} \]

First calculate fraction:

\[ \frac{0.0025}{0.0005}=5 \]

Thus

\[ C=8.854\times10^{-12}\times5 \] \[ C=44.27\times10^{-12}F \] \[ C=44.27\,pF \]

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Step 3: Relation Between Displacement and Capacitance Change

Differentiate capacitance equation:

\[ C=\frac{\varepsilon A}{d} \] \[ \frac{dC}{dd}=-\frac{\varepsilon A}{d^2} \]

Magnitude of sensitivity:

\[ S=\frac{\varepsilon A}{d^2} \]

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Step 4: Substitute Values

\[ S=\frac{8.854\times10^{-12}\times0.0025}{(5\times10^{-4})^2} \]

Square the denominator:

\[ (5\times10^{-4})^2=25\times10^{-8}=2.5\times10^{-7} \]

Now numerator:

\[ 8.854\times10^{-12}\times0.0025=2.2135\times10^{-14} \]

Thus

\[ S=\frac{2.2135\times10^{-14}}{2.5\times10^{-7}} \] \[ S=8.85\times10^{-8} F/m \]

Convert to pF/mm

\[ 1F=10^{12}pF \] \[ S=88.5\,pF/mm \]

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Final Result

\[ S \approx 88.5\,pF/mm \]

Closest option in MCQ:

B. 66.67 pF/mm