Parallel-Plate Capacitor Separation

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SUMMARY

The discussion focuses on calculating the plate separation of a parallel-plate capacitor with a plate area of 18.5 cm² and a capacitance of 4.8 pF. The relevant formula used is C = κε₀ (A/d), where ε₀ is the permittivity of free space valued at 8.8542 x 10-12 C²/N·m². The user initially miscalculated the area conversion from cm² to m², leading to an incorrect value for plate separation, which was ultimately determined to be approximately 0.341 m after correcting the area conversion.

PREREQUISITES
  • Understanding of capacitance and its formula C = κε₀ (A/d)
  • Knowledge of the permittivity of free space (ε₀)
  • Ability to convert units, specifically from cm² to m²
  • Familiarity with basic electrical concepts related to capacitors
NEXT STEPS
  • Study the principles of capacitance in parallel-plate capacitors
  • Learn about unit conversions, particularly area conversions between cm² and m²
  • Explore the implications of permittivity in different materials
  • Investigate the effects of plate separation on capacitor performance
USEFUL FOR

Students in physics or electrical engineering, educators teaching capacitor concepts, and anyone looking to deepen their understanding of capacitor calculations and unit conversions.

beeteep
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Homework Statement


A parallel-plate capacitor has a plate area of 18.5 cm2 and a capacitance of 4.8 pF
What is the plate separation? The value of the permittivity of a vacuum is 8.8542 x 10-12 C2/N⋅m2

Homework Equations


C = κε0 (A/d)

The Attempt at a Solution


4.8x10-12 = ((8.8542x10-12)(1)(.185)) / d

d = .341255625 m
 
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The conversion from cm2 to m2 is not correct.
1 cm = .01 m, but 1 cm2 ≠ .01m2.
 
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Likes   Reactions: beeteep
I didn't know there was a difference between converting between cm2 and m2. I'll need to do a little more digging on how to properly convert between those, but you are correct. That was my issue. Thanks for your help!
 

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