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Calculate the area of the plates that will produce a 3F capacitor if it is desired to separate the plate by a 6 mm air gap.

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Answer

2.034×109 m2\text{2.034} \times 10^9 \text{ m}^2

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Here's the solution for .5:

1.5) Calculate the area of the plates that will produce a 3F capacitor if it is desired to separate the plate by a 6 mm air gap.

Given values:
Capacitance, C=3 FC = 3 \text{ F}
Plate separation, d=6 mm=6×10−3 md = 6 \text{ mm} = 6 \times 10^{-3} \text{ m}
Permittivity of free space (for air gap), ϵ0=8.85×10−12 F/m\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}

  1. Use the formula for the capacitance of a parallel plate capacitor.

    C=ϵ0AdC = \frac{\epsilon_0 A}{d}

    We need to find the area AA, so rearrange the formula:

    A=Cdϵ0A = \frac{C d}{\epsilon_0}

  2. Substitute the given values into the rearranged formula.

    A=(3 F)×(6×10−3 m)8.85×10−12 F/mA = \frac{(3 \text{ F}) \times (6 \times 10^{-3} \text{ m})}{8.85 \times 10^{-12} \text{ F/m}}

  3. Calculate the area.

    A=18×10−3 F⋅m8.85×10−12 F/mA = \frac{18 \times 10^{-3} \text{ F}\cdot\text{m}}{8.85 \times 10^{-12} \text{ F/m}} A≈2.034×109 m2A \approx 2.034 \times 10^9 \text{ m}^2

    The area of the plates required is 2.034×109 m2\text{2.034} \times 10^9 \text{ m}^2.

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