Electric field problem - Plastic Slab

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SUMMARY

The discussion centers on calculating the electric field and potential due to a plastic slab with a uniform free charge density (+rho) and a surface charge (-sigma). The user applied Gauss's Law, resulting in the equation E = ((rho)(t) - sigma) / 2epsilon for the electric field outside the slab. However, the user overlooked that the electric field inside the slab varies with distance from the edges, which is crucial for accurately plotting E as a function of distance (r). The need to differentiate between the electric fields inside and outside the slab is emphasized for a complete solution.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric fields and charge distributions
  • Knowledge of the concepts of free charge density and surface charge density
  • Basic skills in sketching functions and interpreting graphs
NEXT STEPS
  • Study the application of Gauss's Law in different geometries
  • Learn about electric fields within dielectric materials
  • Explore the concept of electric potential and its relationship to electric fields
  • Investigate how to plot electric fields as functions of distance in various charge configurations
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Students and professionals in physics, particularly those studying electromagnetism, electrical engineers, and anyone interested in understanding electric fields in charged materials.

meteorologist1
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Please consider this problem:
A plastic slab of thickness t has a uniform free charge density, +rho, embedded inside, and also one surface has a surface charge of -sigma. Find the electric fields here and sketch as a function of distance from one surface. Also find the potential as a function of distance, sketch results. [Ignore issues of dielectric polarizability, the function of the plastic is simply to fix the charges in place]

Here is what I did: I draw a Gaussian pillbox with area A. Only the top and bottom caps contribute where E and dA are parallel. The charge enclosed is: (rho)(A)(t) + (-sigma)(A) = A((rho)(t) - sigma)

By Gauss's Law, we have: 2EA = [A((rho)(t) - sigma)] / epsilon, so therefore, E = ((rho)(t) - sigma) / 2epsilon

I'm not sure what I did wrong here since the question asks to plot E as a function of r. In my case I would have a constant line. What did I do wrong? Thanks.
 
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Maybe I also have to find the E field within the slab... That answer would depend on r. Any thoughts?
 
meteorologist1 said:
Maybe I also have to find the E field within the slab... That answer would depend on r. Any thoughts?

Yes. You've found the field outside the slab. The field inside depends on distance from edge.
 

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