Electric Field Strenght on a point due to surface charge from an infinte sheet

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SUMMARY

The discussion focuses on calculating the electric field strength due to a non-conducting infinite sheet with a surface charge density of 0.3540 µC/m². The relevant formula for the electric field (E) from an infinite sheet is E = σ / (2ε₀), where ε₀ is the permittivity of free space. The distance from the sheet does not affect the electric field strength for points close to the sheet, as the field remains approximately constant. Therefore, the electric field at a distance of 6.40 cm from the sheet is calculated using the surface charge density without needing to incorporate the distance into the formula.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with the concept of surface charge density
  • Knowledge of the permittivity of free space (ε₀)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the electric field equation for an infinite sheet
  • Learn about the implications of surface charge density on electric fields
  • Explore the concept of electric field uniformity in relation to distance from charged surfaces
  • Investigate applications of electric fields in real-world scenarios, such as capacitors
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Students in physics, particularly those studying electromagnetism, educators teaching electric field concepts, and anyone interested in understanding the behavior of electric fields generated by charged surfaces.

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Homework Statement


A large, thin, non-conducting sheet has a surface charge density of 0.3540 [tex]\mu[/tex]C/m[tex]^{2}[/tex]. Determine the magnitude of the electric field at a distance of 6.40 cm from the sheet.


Homework Equations


Electric Field of Charged Disc
E=[tex]\frac{\sigma}{2e\ =\ 1.602176462(63)\ \times\ 10^{-19}\ C}[/tex](1-[tex]\frac{z}{\sqrt{z^{2}+R^{2}}}[/tex])

Infinite sheet
E=[tex]\frac{\sigma}{2e}[/tex]


The Attempt at a Solution


I have no idea what to do here. I assume I use the infinite sheet equation but how would I use the 6.4 cm to find the field strength there? I feel like maybe I have to use the equation for field strength E=[tex]\frac{kQ}{r^{2}}[/tex] some how but I have no ideas.


Sorry for the sloppy post this is my first post on these boards...Help please.
 
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Failbot said:

The Attempt at a Solution


I have no idea what to do here...
Infinite sheet
E=[tex]\frac{\sigma}{2e}[/tex]

If you have a formula you believe is correct, why not use it? Sure, you can't substitute the distance into the formula, but that is because the formula says that the field is approximately constant for distances small compared to the expanse of the slab, so distance does not matter.

An analogous question in mechanics would be, "what is the gravitational field 1 meter off the ground of the Earth?". Then someone else might ask, "what is the gravitational field 100 meters off the ground of the Earth?". Hopefully the fact that someone threw in the exact height, even though it is not critical, would not prevent you from answering the question.
 

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