Electric field due to distributed planar charge

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SUMMARY

The discussion focuses on deriving the electric field due to a distributed planar charge on an annulus with inner radius R1 and outer radius R2, positioned in the xz plane. The surface charge density is denoted as σ, and the electric field E is calculated at a distance Y above the annulus along the +Y-axis. The derived expression for the electric field in the y direction is E_y = Kλ2πy[-1/(r^2+y^2)^(1/2) + 1/(r^2+y^2)^(1/2)], indicating the use of cylindrical coordinates for integration from r1 to r2 to account for the annulus' geometry.

PREREQUISITES
  • Understanding of electric fields and Coulomb's law
  • Familiarity with surface charge density concepts
  • Knowledge of cylindrical coordinate systems
  • Basic integration techniques in calculus
NEXT STEPS
  • Study the derivation of electric fields from continuous charge distributions
  • Learn about the application of cylindrical coordinates in electrostatics
  • Explore integration techniques for calculating fields from annular charge distributions
  • Investigate the implications of varying surface charge densities on electric field calculations
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism, as well as engineers and researchers working with electric fields and charge distributions.

easybakeoven
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An annulus (disk with a concentric hole) has inner and outer radii of R1 and R2 respectively, and uniform surface charge density of \sigma. the annulus lies in the xz plane with the y-axis centered in the hole.

a.) using the most basic expression of point charge electric field as a starting point, derive an expression for the electric field E at some distance Y above the annulus along the +Y-axis.


Homework Equations



E=kQ/r^2


The Attempt at a Solution



I started working on the problem, and I think that the equation I've come down to for the electric field in the y direction may be close to right...

E_y=K\lambda2\piy*[-1/(r^2+y^2)^(1/2) + 1/(r^2+y^2)^(1/2)]

Is this close to right? I tried to work the problem by looking at a small linear slice of the annulus, and then rotating that around the y-axis
 
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I guess you could do that though I am not sure of the math involved (I think you would use the cylindrical coordinate system to make it easier), Id find the electric field due to a ring of radius dr and then integrate that expression from r1 to r2.
 

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