Nonconducting surface uniform charge hole

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SUMMARY

The discussion focuses on calculating the electric field at a point P along the axis of a circular hole cut into a large, flat, nonconducting surface with a uniform charge density σ. The electric field from the surface is given by E1 = +σ/(2ε0), while the electric field from the hole, treated as a disc with a uniform charge density of -σ, is E2 = -σ/(2ε0)(1 - z/√(z2 + R2)). The net electric field at point P is determined by combining these two fields.

PREREQUISITES
  • Understanding of electric fields from charged surfaces and discs
  • Knowledge of the concept of uniform charge density
  • Familiarity with the permittivity of free space (ε0)
  • Basic algebra for manipulating equations involving electric fields
NEXT STEPS
  • Study the derivation of electric fields from infinite charged planes
  • Learn about the superposition principle in electrostatics
  • Explore the concept of fringing fields and their effects on electric field calculations
  • Investigate applications of electric fields in capacitors and other electronic components
USEFUL FOR

Students studying electrostatics, physics educators, and anyone interested in understanding electric fields in nonconducting materials.

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Homework Statement


Didn't know what to put for thread title, my notes and textbook were a little to ambiguous

A large, flat, nonconducting surface carries a uniform charge density σ. Into the middle of this sheet has been cut a small, circular hole of radius R. Ignoring fringing fields from the edges, calculate the electric field at a point, P that is a distance z from the center of the hole along its axis. hint: consider the electric field from a sheet of charge and the electric field from a disk of charge
edit** nevermind i was being dumb, i got it
 
Last edited:
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You can consider the large, flat, nonconducting surface carries a uniform charge density +σ

and the hole as a disc carrying a uniform charge density -σ.

The electric field due to the surface is

[tex]E_1 = \frac{+\sigma}{2\epsilon_{0}}[/tex]

The electric field due to the disc is

[tex]E_2 = \frac{-\sigma}{2\epsilon_{0}}(1 - \frac{z}{\sqrt{z^2 + r^2}})[/tex]

Now find the net field at P.
 
rl.bhat said:
You can consider the large, flat, nonconducting surface carries a uniform charge density +σ

and the hole as a disc carrying a uniform charge density -σ.

The electric field due to the surface is

[tex]E_1 = \frac{+\sigma}{2\epsilon_{0}}[/tex]

The electric field due to the disc is

[tex]E_2 = \frac{-\sigma}{2\epsilon_{0}}(1 - \frac{z}{\sqrt{z^2 + r^2}})[/tex]

Now find the net field at P.

that is exactly how i did it. thanks, i finished it yesterday, but I'm not sure how to delete threads
 

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