Anonymous
Jan 11
The area of one plate of a two plate mica capacitor is 0.0025m² and the separation between plates is 0.02m. if the dielectrics constant of mica is 7. Find the capacitance of the capacitor?
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Jan 11
The capacitance of a capacitor can be calculated using the formula:
C = (ε₀ * εᵣ * A) / d
where:
C is the capacitance
ε₀ is the permittivity of free space (8.854 x 10⁻¹² F/m)
εᵣ is the relative permittivity (dielectric constant) of the material (7 for mica)
A is the area of one plate (0.0025 m²)
d is the separation between plates (0.02 m)
Plugging in the values:
C = (8.854 x 10⁻¹² F/m * 7 * 0.0025 m²) / 0.02 m
C = (8.854 x 10⁻¹² F/m * 7 * 0.0025 m²) / 0.02 m
C = 0.001239 F
Therefore, the capacitance of the capacitor is approximately 0.001239 Farads.
C = (ε₀ * εᵣ * A) / d
where:
C is the capacitance
ε₀ is the permittivity of free space (8.854 x 10⁻¹² F/m)
εᵣ is the relative permittivity (dielectric constant) of the material (7 for mica)
A is the area of one plate (0.0025 m²)
d is the separation between plates (0.02 m)
Plugging in the values:
C = (8.854 x 10⁻¹² F/m * 7 * 0.0025 m²) / 0.02 m
C = (8.854 x 10⁻¹² F/m * 7 * 0.0025 m²) / 0.02 m
C = 0.001239 F
Therefore, the capacitance of the capacitor is approximately 0.001239 Farads.
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