Calculating Surface Charge Density in a Uniform Electric Field

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SUMMARY

The discussion centers on calculating the surface charge density (\(\sigma\)) on two large conducting plates separated by 2.3 cm in a uniform electric field. The formula derived is \(\sigma = \frac{mg\epsilon_{0}}{q}\), where \(m\) is the mass of the electron, \(g\) is the acceleration due to gravity, \(q\) is the charge of the electron, and \(\epsilon_{0}\) is the permittivity of free space. Participants debated the relevance of the distance between the plates, concluding it does not affect the surface charge density calculation, which relies on the infinite-plate approximation.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with the concepts of surface charge density
  • Knowledge of Gauss's Law
  • Basic physics principles, including gravitational force
NEXT STEPS
  • Study the application of Gauss's Law in electrostatics
  • Explore the concept of infinite plate approximation in electric fields
  • Learn about the relationship between charge, mass, and electric fields
  • Investigate the role of permittivity of free space (\(\epsilon_{0}\)) in electrostatics
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Physics students, electrical engineers, and anyone interested in electrostatics and electric field calculations.

stunner5000pt
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An electron remains stationary in na electric field directed downward in the Earth's gravitational field. If the electric field is due to charge on the two large conducting plates oppositely charged and separated by 2.3cm what is the surface charge density assumed to be uniform on the plates?

force on the elctric due to the plates is qE
and E = \sigma/epsilon_{0}
force of gfravity is mg
so \sigma = \frac{mg\epsilon_{0}}{q}
and this is the surface charge density on the plates, isn't it ?

Does the distnace (which does NOT appear on the formula and is totally irrelevant i think) matter?
Please do advise on this matter!
Thank you!
 
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I say distance doesn't matter.

Maybe, just maybe, the distance was given to convince you that the infinite-plate approximation was justified.

Btw - the thread title advertises Gauss law but I don't see none. I was ready for some good ol' Gauss action here. kinda disapointed :P
 
Last edited:

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