What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of k=5?
Practice Questions
1 question
Q1
What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of k=5?
5.5 µF
5.0 µF
4.5 µF
6.0 µF
C = k * ε0 * (A/d) = 5 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m) = 5.0 µF.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the capacitance of a parallel plate capacitor with an area of 0.01 m² and a separation of 0.001 m, filled with a dielectric of k=5?
Solution: C = k * ε0 * (A/d) = 5 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m) = 5.0 µF.
Steps: 7
Step 1: Identify the formula for the capacitance (C) of a parallel plate capacitor: C = k * ε0 * (A/d).
Step 2: Determine the values needed for the formula: Area (A) = 0.01 m², separation (d) = 0.001 m, dielectric constant (k) = 5, and ε0 (the permittivity of free space) = 8.85 x 10^-12 F/m.
Step 3: Plug the values into the formula: C = 5 * (8.85 x 10^-12 F/m) * (0.01 m² / 0.001 m).
Step 4: Calculate the area divided by the separation: 0.01 m² / 0.001 m = 10.
Step 5: Multiply the values: C = 5 * (8.85 x 10^-12 F/m) * 10.
Step 6: Calculate the final capacitance: C = 5 * 8.85 x 10^-11 F = 4.425 x 10^-11 F.
Step 7: Convert the capacitance to microfarads (µF): 4.425 x 10^-11 F = 4.425 x 10^-5 µF = 5.0 µF (rounded).
