To calculate the effective plate area of a capacitor, we will use the capacitance formula for a parallel plate capacitor with a dielectric material:
[tex]\[ C = \frac{\epsilon_r \cdot \epsilon_0 \cdot A}{d} \][/tex]
Where:
- [tex]\( C \)[/tex] is the capacitance
- [tex]\( \epsilon_r \)[/tex] is the relative permittivity of the dielectric
- [tex]\( \epsilon_0 \)[/tex] is the permittivity of free space ([tex]\( \epsilon_0 = 8.854 \times 10^{-12} \, F/m \)[/tex])
- [tex]\( A \)[/tex] is the area of one of the plates
- [tex]\( d \)[/tex] is the thickness of the dielectric
Given data:
- Capacitance, [tex]\( C = 50 \times 10^{-6} \, F \)[/tex]
- Relative permittivity, [tex]\( \epsilon_r = 1.4 \)[/tex]
- Thickness of the dielectric, [tex]\( d = 0.2 \times 10^{-6} \, m \)[/tex]
We need to rearrange the capacitance formula to solve for the plate area [tex]\( A \)[/tex]:
[tex]\[ A = \frac{C \cdot d}{\epsilon_r \cdot \epsilon_0} \][/tex]
Substituting the given values into the formula, we get:
[tex]\[ A = \frac{50 \times 10^{-6} \, F \times 0.2 \times 10^{-6} \, m}{1.4 \times 8.854 \times 10^{-12} \, F/m} \][/tex]
Now, let's perform the calculation step-by-step.
1. Multiply the capacitance [tex]\( C \)[/tex] by the thickness [tex]\( d \)[/tex]:
[tex]\[ 50 \times 10^{-6} \, F \times 0.2 \times 10^{-6} \, m = 10 \times 10^{-12} \, F \cdot m \][/tex]
2. Multiply the relative permittivity [tex]\( \epsilon_r \)[/tex] by the permittivity of free space [tex]\( \epsilon_0 \)[/tex]:
[tex]\[ 1.4 \times 8.854 \times 10^{-12} \, F/m = 12.3956 \times 10^{-12} \, F/m \][/tex]
3. Divide the result from step 1 by the result from step 2:
[tex]\[ \frac{10 \times 10^{-12} \, F \cdot m}{12.3956 \times 10^{-12} \, F/m} \approx 0.8067378747297429 \, m^2 \][/tex]
Therefore, the effective plate area of the capacitor is approximately [tex]\( 0.807 \, m^2 \)[/tex].
