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You need to construct a 400 pF capacitor for a science project. You plan to cut two L x L metal squares and place spacers between them. The thinnest spacers you have are 0.20 mm thick. What is the proper value of L?Express your answer in centimeters.

User Dvska
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2 Answers

14 votes
14 votes

Final answer:

To determine the proper value of L for the metal squares in order to construct a 400 pF capacitor, use the capacitance formula for a parallel-plate capacitor and solve for L.

Step-by-step explanation:

The question pertains to constructing a parallel-plate capacitor with a specified capacitance of 400 pF using metal plates with spacers in between them. To determine the proper value of L for the metal squares, we can use the formula for the capacitance of a parallel-plate capacitor:

C = ε₀(εr)(A/d)

Where ε₀ is the permittivity of free space, εr is the relative permittivity or dielectric constant (for air, εr is close to 1), A is the area of the plates in square meters, and d is the separation between the plates in meters. Given the capacitance (C = 400 pF), the thickness of the spacers (d = 0.20 mm), and that we're using air as the dielectric as the spacers are thin:

C = (8.85 × 10⁻¹² F/m)(1)(L²/0.20 × 10⁻³ m)

Solving for L and converting it to centimeters gives:

L = √((400 × 10⁻±² F)(0.20 × 10⁻³ m)/(8.85 × 10⁻¹² F/m)) × 100 cm/m

This calculation will yield the proper value of L in centimeters.

User FoxyBOA
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29 votes
29 votes

Consider that the formula for the capacitance of a square parallel plate capacitor is:


C=\epsilon_o(A)/(d)=\epsilon_o(L^2)/(d)

where A=L^2 is the area of each plate, d is the separation between plates and

ε0 is the dielectric permitivity of vacuum ans is equal to 8.82*10^-12 F/m.

If you solve the previous expression for L and replace the given values for d and C, you obtain:


\begin{gathered} L=\sqrt[]{(dC)/(\epsilon_o)} \\ d=0.20mm=0.20\cdot10^(-3)m=2.0\cdot10^(-4)m \\ C=400pF=400\cdot10^(-12)F=4.00\cdot10^(-10)F \\ L=\sqrt[]{((2.0\cdot10^(-4)m)(4.00\cdot10^(-10)F))/(8.85\cdot10^(-10)(F)/(m))} \\ L\approx0.0095m=0.95cm \end{gathered}

Hence, the proper value of L to construct the required capacitor is approximately 0.95cm

User Albatross
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