What is the capacitance of this capacitor? The plates on a vacuum capacitor have a radius of 2.5 mm and are separated by a distance of 0.75 mm.
O 2.3 x 10-13 F
O 9.3 X 10-11 F
O 3.0 10-11 F
O 2.3 * 10-10 F
Question
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Answer:
2.3 x 10-13 F
Explanation:
From the question,
Applying
C = e₁e₂A/d………………… Equation 1
Where C = Capacitance of the capacitor, d = distance of seperation, A = cross sectional area of the capacitor, e₁ = permitivity of free space, e₂ = relative permitivity of vacuum.
But,
Area of a circle (A) is given as
A = πr²………………. Equation 2
Where r = radius of the capacitor.
Substitute equation 2 into equation 1
C = e₁e₂πr²/d…………… Equation 3
Given: r = 2.5 mm = 0.0025 m, d = 0.75 mm = 0.00075 m
Constant; e₁ = 8.85×10⁻¹² F/m, e₂ = 1 F/m, π = 3.14.
Substitute these values into equation 3
C = [(8.85×10⁻¹²)(1)( 0.0025²)(3.14)]/(0.00075)
C = 0.23×10⁻¹²
C = 2.3×10⁻¹³ F