How Do You Calculate the Potential Difference in a Parallel Plate Capacitor?

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SUMMARY

The discussion focuses on calculating the potential difference across a parallel plate capacitor with an area of 4.77 cm², a plate separation of 1.06 mm, and a charge of 403 pC. The correct conversion of area to square meters is 0.000477 m², and the distance should be converted to 0.00106 m. The dielectric constant used is 1.00054. The formula applied is V = Q / (ε₀ * A / d), leading to a calculated potential difference of 1.01 V, which was identified as incorrect due to an error in area conversion.

PREREQUISITES
  • Understanding of capacitor fundamentals
  • Knowledge of the formula for potential difference in capacitors
  • Familiarity with unit conversions (cm² to m², mm to m)
  • Basic grasp of dielectric constants
NEXT STEPS
  • Review the formula for potential difference in capacitors: V = Q / (ε * A / d)
  • Practice unit conversions specifically for area and distance measurements
  • Explore the impact of different dielectric materials on capacitance
  • Learn about the implications of charge storage in capacitors
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Students in physics or electrical engineering, educators teaching capacitor concepts, and anyone involved in electronics design or analysis.

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A parallel-plate capacitor has an area of 4.77 cm2, and the plates are separated by 1.06 mm with air between them. It stores a charge of 403 pC. What is the potential difference across the plates of the capacitor?

alright so i changed the area to .0477m^2. then i changed the distance between the plates to .00106m. i used 1.00054 as the dielectric constant. and i changed the charge to 403E-12C.

then i took 403E-12=((1.00054)(8.85E-12)(.0477)/(.00106))V

and i get V to equal 1.01V and its wrong. any help is appreciated.
 
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For one thing, 4.77 cm^2 is 0.000477 m^2.
 

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