Electric Field Above Square: Find E-Field at z Height

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SUMMARY

The discussion focuses on calculating the electric field at a height z above a square of side length a with a uniform surface charge density σ. The integral used to derive the electric field is given as E = (zσ / (4πε₀)) ∫∫ (dx dy) / (x² + y² + z²)^(3/2), which is intended to be evaluated over the limits from -a/2 to a/2. Participants identify that the integral setup may contain errors, leading to incorrect results, and seek clarification on the proper formulation.

PREREQUISITES
  • Understanding of electric fields and surface charge density
  • Familiarity with double integrals in multivariable calculus
  • Knowledge of the constants ε₀ (permittivity of free space) and π
  • Experience with vector calculus and integration techniques
NEXT STEPS
  • Review the derivation of electric fields from surface charge distributions
  • Study the application of double integrals in calculating fields
  • Explore the concept of electric field lines and their properties
  • Investigate common pitfalls in integral setups for electrostatics problems
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Students in physics, particularly those studying electromagnetism, as well as educators and anyone involved in solving electrostatics problems related to charge distributions.

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Homework Statement


Find the electric field a height z above a square of side length [tex]a[/tex] with uniform surface charge [tex]\sigma[/tex]

Homework Equations


The Attempt at a Solution


[tex]E=\frac{z\sigma}{4\pi\epsilon_0}\int^{a/2}_{-a/2}\int^{a/2}_{-a/2}\frac{dx dy}{(x^2+y^2+z^2)^{3/2}}[/tex]
(in the z direction). However, unless I maple'd wrong, this gives the wrong answer. Can you anyone see a problem with the way the integral is set up?
 
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