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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​

2 Answers

4 votes

Answer:

A

Step-by-step explanation:

edge

User Ryan Fitzgerald
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5 votes

Answer:

2.3 x 10-13 F

Step-by-step 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

User Vanomart
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