If the area of each plate of a parallel plate capacitor is 0.01 m², the distance between the plates is 1 mm, and the relative permittivity of the dielectric is 3.0, what is the cap... If the area of each plate of a parallel plate capacitor is 0.01 m², the distance between the plates is 1 mm, and the relative permittivity of the dielectric is 3.0, what is the capacitance of the capacitor?
Understand the Problem
The question is asking to calculate the capacitance of a parallel plate capacitor given the area of the plates, the distance between the plates, and the relative permittivity of the dielectric material.
Answer
The capacitance is \( C = 0.2655 \, \mu\text{F} \).
Answer for screen readers
The capacitance of the capacitor is ( C = 0.2655 , \mu\text{F} ).
Steps to Solve
- Identify the formula for capacitance To calculate the capacitance ($C$) of a parallel plate capacitor, use the formula: $$ C = \frac{\varepsilon_r \varepsilon_0 A}{d} $$
Where:
- $C$ = capacitance
- $\varepsilon_r$ = relative permittivity of the dielectric
- $\varepsilon_0$ = permittivity of free space ($8.85 \times 10^{-12} , \text{F/m}$)
- $A$ = area of one of the plates
- $d$ = distance between the plates
- Substitute the known values into the formula Given:
- Area $A = 0.01 , \text{m}^2$
- Distance $d = 1 , \text{mm} = 0.001 , \text{m}$
- Relative permittivity $\varepsilon_r = 3.0$
Now substitute these values into the capacitance formula: $$ C = \frac{3.0 \times 8.85 \times 10^{-12} , \text{F/m} \times 0.01 , \text{m}^2}{0.001 , \text{m}} $$
- Calculate the capacitance First, calculate the numerator: $$ 3.0 \times 8.85 \times 10^{-12} \times 0.01 = 2.655 \times 10^{-13} , \text{F} $$
Next, calculate the capacitance: $$ C = \frac{2.655 \times 10^{-13} , \text{F}}{0.001 , \text{m}} = 2.655 \times 10^{-10} , \text{F} $$
- Convert to microfarads To express the final answer in microfarads (1 µF = $10^{-6}$ F): $$ C = 2.655 \times 10^{-10} , \text{F} = 0.2655 , \mu\text{F} $$
The capacitance of the capacitor is ( C = 0.2655 , \mu\text{F} ).
More Information
A parallel plate capacitor stores electrical energy and its capacitance depends on the area of the plates, the distance between them, and the dielectric material used. The values calculated here lead to a practical capacitor that can be used in various electronic applications.
Tips
- Incorrect unit conversion: Ensure the distance is converted from mm to m properly.
- Forgetting to use the permittivity of free space: Always include $\varepsilon_0$ in calculations as it’s essential for determining capacitance.
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