Charge between two metal plates

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SUMMARY

The discussion focuses on calculating the charge |q| between two large metal plates with an area of 1.1 m², separated by 6.2 cm, and an electric field magnitude E of 50 N/C. The relevant equations used include |q|=σA and E=σ/(2ε₀), leading to the formula |q|=2ε₀EA. The calculated charge is |q|=9.889E-7 C, derived using the permittivity of free space ε₀ as 8.99E-9 C/N·m². A correction was noted regarding the use of the permittivity of free space instead of Coulomb's constant.

PREREQUISITES
  • Understanding of electric fields and charge distribution
  • Familiarity with the concept of permittivity of free space (ε₀)
  • Knowledge of basic electrostatics equations
  • Ability to manipulate algebraic equations for physics problems
NEXT STEPS
  • Study the derivation of the electric field between parallel plates
  • Learn about the implications of fringing effects in electric fields
  • Explore the relationship between charge density (σ) and electric field (E)
  • Investigate applications of electrostatics in capacitor design
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding electrostatics, particularly in the context of parallel plate capacitors and electric fields.

Jrlinton
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Homework Statement


Two large metal plates of area 1.1 m^2 face each other, 6.2 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 50 N/C. Find |q|.

Homework Equations


|q|=σA
E=σ/(2Eo)⇒σ=2EoE
|q|=2EoEA

The Attempt at a Solution


I thought that after I did the rearranging of formulas above that this would be a pretty simple problem
|q|=2EoEA
|q|=2(8.99E-9C)(50N/C)(1.1m^2)
|q|=9.889E-7C
 
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Jrlinton said:

Homework Statement


Two large metal plates of area 1.1 m^2 face each other, 6.2 cm apart, with equal charge magnitudes but opposite signs. The field magnitude E between them (neglect fringing) is 50 N/C. Find |q|.

Homework Equations


|q|=σA
E=σ/(2Eo)⇒σ=2EoE
|q|=2EoEA

The Attempt at a Solution


I thought that after I did the rearranging of formulas above that this would be a pretty simple problem
|q|=2EoEA
|q|=2(8.99E-9C)(50N/C)(1.1m^2)
|q|=9.889E-7C
There are two plates.
 
It seems that I used coulombs constant instead of the permittivity of free space
 

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