Is the Electric Field Calculation for an Infinite Sheet Correct?

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SUMMARY

The discussion centers on the calculation of the electric field generated by an infinite sheet of charge using the Coulomb rule. The user outlines their approach, which includes deriving expressions for the electric field components and setting up the integral for evaluation. The initial steps involve determining the relationships between variables such as radius (r), angle (Theta), and charge density (Sigma). The consensus confirms that the user's methodology is correct up to the point of setting the integral limits from Theta = 0 to Theta = 90 degrees.

PREREQUISITES
  • Coulomb's Law for electric fields
  • Basic calculus for integration
  • Trigonometric functions and their applications
  • Understanding of electric field concepts in electrostatics
NEXT STEPS
  • Calculate the integral for the electric field using the derived expressions
  • Explore the implications of the electric field result for infinite sheets of charge
  • Review the concept of charge density (Sigma) and its impact on electric fields
  • Study the behavior of electric fields in different geometries beyond infinite sheets
USEFUL FOR

Students and educators in physics, particularly those focusing on electrostatics and electric field calculations, as well as anyone involved in advanced physics problem-solving.

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Homework Statement



What is the field?
I just want you to chek my solution (just the begining) (math)

You can see the paint.

Homework Equations



Coulomb rule

The Attempt at a Solution


1.
tan(Tetha) = r/a
=> r = a*tan(Tetha)
=> dr = a/cos^2(Tetha)d(Tetha)
2.
dq = Sigma * da = 2Pi*Sigma*rdr =
= 2Pi*Sigma*r*a/cos^2(Tetha)d(Tetha)
3.
cos(Tetha) = a/R
=>1/R^2 = cos^2(Tetha)/a^2
4.
dEy (the enqual) = dE * cos(Tetha)
5.
The limits of the integral are... from Tetha = 0 to Tetha = 90 deg (?)

I ask you... am I right?

Thanks for the watchers :)
 

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Yes, you are correct so far. The next step would be to calculate the integral, using the equations and limits you have provided. The result of the integral will be the field.
 

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