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What is the capacitance of a parallel plate capacitor with an area of 0.01 m² an

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Question: What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of relative permittivity 5?

Options:

  1. 5.5 × 10^-11 F
  2. 5.5 × 10^-10 F
  3. 5.5 × 10^-9 F
  4. 5.5 × 10^-8 F

Correct Answer: 5.5 × 10^-10 F

Solution: 

C = ε₀ * ε_r * A / d = (8.85 × 10^-12 F/m) * 5 * (0.01 m²) / (0.001 m) = 4.425 × 10^-10 F.

What is the capacitance of a parallel plate capacitor with an area of 0.01 m² an

Practice Questions

Q1
What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of relative permittivity 5?
  1. 5.5 × 10^-11 F
  2. 5.5 × 10^-10 F
  3. 5.5 × 10^-9 F
  4. 5.5 × 10^-8 F

Questions & Step-by-Step Solutions

What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of relative permittivity 5?
  • Step 1: Identify the formula for the capacitance (C) of a parallel plate capacitor: C = ε₀ * ε_r * A / d.
  • Step 2: Determine the values needed for the formula: ε₀ (the permittivity of free space) is approximately 8.85 × 10^-12 F/m, ε_r (the relative permittivity of the dielectric) is 5, A (the area of the plates) is 0.01 m², and d (the separation between the plates) is 0.001 m.
  • Step 3: Substitute the values into the formula: C = (8.85 × 10^-12 F/m) * 5 * (0.01 m²) / (0.001 m).
  • Step 4: Calculate the numerator: (8.85 × 10^-12 F/m) * 5 * (0.01 m²) = 4.425 × 10^-13 F/m.
  • Step 5: Divide the result from Step 4 by d (0.001 m): 4.425 × 10^-13 F/m / 0.001 m = 4.425 × 10^-10 F.
  • Step 6: Conclude that the capacitance of the capacitor is 4.425 × 10^-10 F.
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