Electric Field due to a charged conducting sheet

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SUMMARY

The discussion centers on calculating the electric field magnitude due to a charged conducting sheet measuring 10m x 10m x 1mm with a charge of 1 mC. The correct formula for electric field strength is E = σ/2ε0, where σ is the surface charge density. The user initially calculated the volume charge density incorrectly, leading to confusion in the final result. The correct electric field magnitude is 565 kV/m, achieved by using the appropriate surface charge density and the correct value for free space permittivity, ε0 = 8.85 x 10^-12 F/m.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with the concept of surface charge density (σ)
  • Knowledge of free space permittivity (ε0)
  • Basic proficiency in physics equations related to electrostatics
NEXT STEPS
  • Study the derivation and applications of the formula E = σ/2ε0
  • Learn about the differences between volume charge density and surface charge density
  • Explore the implications of free space permittivity in electric field calculations
  • Investigate real-world applications of electric fields in conducting materials
USEFUL FOR

Students studying electromagnetism, physics educators, and professionals in electrical engineering seeking to deepen their understanding of electric fields and charge distributions in conductive materials.

vodkasoup
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Homework Statement



A conducting sheet 10m x 10m x 1mm carries a charge of 1 mC. What is the electric field magnitude in kV m-1 just above the centre of the sheet?

Homework Equations



ρ=Q/Volume

E=ρ/2ε0

The Attempt at a Solution



I have arrived at an answer by the following:

ρ = 1x10-3C / 10m x 10m x 1mm = 0.01C/m3

E = 0.01 / 2 x 8.85x10-12 = 565MV/m

However, the answer is given as 565kV/m, which I can only arrive at if I use 8.85x10-9 as the free space permittivity. Where have I gone wrong?Many thanks.
 
Last edited:
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vodkasoup said:
ρ=Q/Volume

E=ρ/2ε0
The density you need is the surface charge density, Q/Area (which is usually represented by σ). (Not the volume charge density ρ.)
 
Ah. Whoops. Thank you very much!
 

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