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You are asked to design a parallel-plate capacitor having a capacitance of 1.00 f and a plate separation of 1.00 mm. Calculate the required surface area of each plate. is this a realistic size for a capacitor



Answer :

audreyn06

Answer: 1.13 · 10⁸ m²; No

Explanation:

The formula for the capacitance of a parallel-plate capacitor is given as:

C = ε₀A/d, where:

  • C is the capacitance (F)
  • ε₀ is the vacuum permittivity (8.85 · 10⁻¹² F/m)
  • A is the surface area of each plate (m²)
  • d is the separation between the plates (m)

We are given:

  • C = 1 F
  • d = 1 mm = 0.001 m

Let's plug our known values into the formula to solve for A:

1 = (8.85 · 10⁻¹²)A/0.001

0.001 = (8.85 · 10⁻¹²)A

A = 1.13 · 10⁸ m²

This is not a realistic size for a capacitor as 1.13 · 10⁸ m² is an extremely large area. It is about the size of a city.

The capacitor requires an area of 1.13 x 10⁸ m² per plate, which is unrealistic.

To design a parallel-plate capacitor having a capacitance of 1.00 F and a plate separation of 1.00 mm required surface area of each plate can be calculated using the formula for the capacitance of a parallel-plate capacitor:

C = (ε₀ * A) / d,

Rearranging this formula to solve for the area A gives as:

A = (C * d) / ε₀

Substituting the known values:

A = (1.00 F * 1.00 x 10⁻³ m) / (8.85 x 10⁻¹² F/m) = 1.13 x 10⁸ m²

This surface area is extremely large, approximately 113 square kilometers, which is not realistic for a practical capacitor design.

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