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The plates of an air-filled parallel-plate capacitor of capacitance 4.90 pf are 2.29 mm apart. What is the area of each plate?

User Tomha
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Final answer:

To determine the area of each plate of a 4.90 pF air-filled parallel-plate capacitor with 2.29 mm plate separation, use the formula A = C × d / ε0 and calculate with C = 4.90 pF and d = 2.29 mm, resulting in A ≈ 1.27×10−4 m2.

Step-by-step explanation:

To find the area of each plate of an air-filled parallel-plate capacitor with a given capacitance, we can use the formula for the capacitance of a parallel-plate capacitor, which is:

C = ε0 (A/d)

Where C is the capacitance, ε0 is the permittivity of free space (8.85×10−12 F/m), A is the area of one plate, and d is the distance between the plates.

Lets rearrange the formula to solve for A:

A = C × d / ε0

Now plug in the given values:

C = 4.90 pF = 4.90×10−12 F
d = 2.29 mm = 2.29×10−3 m

A = (4.90×10−12 F) × (2.29×10−3 m) / (8.85×10−12 F/m)

After calculating, we get:

A ≈ 1.27×10−4 m2

Therefore, the area of each plate is approximately 1.27×10−4 square meters.

User Sify
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