A parallel plate capacitive displacement sensor has a plate area of $2\text{ cm}^2$. The air gap between the plates is decreased by $0.1\text{ mm}$ from an initial value of $0.5\text{ mm}$. The percentage change in the capacitance value is ______ %. (Assume permittivity as $\epsilon_0 = 8.854 \times 10^{-12}\text{ F/m}$)
Capacitive Sensor Change Calculation
The capacitance ($C$) of a parallel plate capacitor is calculated using the formula:
$ C = \frac{\epsilon_0 A}{d} $
where:
- $ \epsilon_0 $ is the permittivity of free space ($8.854 \times 10^{-12}\text{ F/m}$).
- $ A $ is the area of the plates.
- $ d $ is the distance between the plates (air gap).
We are given:
- Plate Area $ A = 2\text{ cm}^2 = 2 \times 10^{-4}\text{ m}^2 $.
- Initial air gap $ d_1 = 0.5\text{ mm} = 0.5 \times 10^{-3}\text{ m} $.
- Decrease in air gap $ \Delta d = 0.1\text{ mm} $.
- Final air gap $ d_2 = d_1 - \Delta d = 0.5\text{ mm} - 0.1\text{ mm} = 0.4\text{ mm} = 0.4 \times 10^{-3}\text{ m} $.
Calculating Capacitance Values
Let $ C_1 $ be the initial capacitance and $ C_2 $ be the final capacitance.
$ C_1 = \frac{\epsilon_0 A}{d_1} $
$ C_2 = \frac{\epsilon_0 A}{d_2} $
Determining Percentage Change
The percentage change in capacitance is given by:
$ \text{Percentage Change} = \frac{C_2 - C_1}{C_1} \times 100\% $
This can be simplified as:
$ \text{Percentage Change} = \left( \frac{C_2}{C_1} - 1 \right) \times 100\% $
Substituting the capacitance formulas:
$ \frac{C_2}{C_1} = \frac{\frac{\epsilon_0 A}{d_2}}{\frac{\epsilon_0 A}{d_1}} = \frac{d_1}{d_2} $
Now, substitute the values of the gaps:
$ \frac{d_1}{d_2} = \frac{0.5\text{ mm}}{0.4\text{ mm}} = \frac{5}{4} = 1.25 $
Calculate the percentage change:
$ \text{Percentage Change} = (1.25 - 1) \times 100\% = 0.25 \times 100\% = 25\% $
The percentage change in the capacitance value is 25%.
Important Questions from Capacitive Transducers
The dynamic characteristics of capacitive transducers are similar to those of
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