Uniform E field for spherical shell.

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SUMMARY

The discussion focuses on determining the constant A for a nonconducting spherical shell with a positive volume charge density ρ = A/r, which ensures a uniform electric field within the shell's thickness (2.07 cm ≤ r ≤ 2.51 cm). The problem involves a central charge of q = 45.8 fC and requires the application of electrostatic principles, specifically superposition and integration to account for the non-constant charge density. The derived expression for the electric field's derivative, ## \frac{Aa^3}{r^4}- \frac{q}{2\pi r^3} ##, indicates the need for further integration to find the total charge before applying Gauss' law.

PREREQUISITES
  • Understanding of electrostatics and electric fields
  • Familiarity with Gauss' law and its applications
  • Knowledge of charge density concepts
  • Ability to perform integration in physics problems
NEXT STEPS
  • Study the application of Gauss' law for non-uniform charge distributions
  • Learn about electric field calculations for spherical charge distributions
  • Explore integration techniques for finding total charge from volume charge density
  • Review the principles of superposition in electrostatics
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Students studying electromagnetism, particularly those tackling problems involving electric fields and charge distributions, as well as educators seeking to enhance their understanding of electrostatic principles.

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Homework Statement


In the figure a nonconducting spherical shell of inner radius a = 2.07 cm and outer radius b = 2.51 cm has (within its thickness) a positive volume charge density ρ = A/r, where A is a constant and r is the distance from the center of the shell. In addition, a small ball of charge q = 45.8 fC is located at that center. What value should A have if the electric field in the shell (arb) is to be uniform?

Homework Equations


## \phi = \frac{q}{\epsilon_{0}} = \oint E \cdot dA ##

The Attempt at a Solution


I found the electric field due to the central charge and ρV at a radius between r (arb), added these together (superposition) and then derived with respect to r, hoping to set the derivative to zero to find my answer, but I got ## \frac{Aa^3}{r^4}- \frac{q}{2\pi r^3} ## as my expression for the derivative, (disregarding epsilon naught, the constant) and obviously I can't set this to zero and disregard r. Not sure where to go from here.
 
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Ok, I can't use Gauss' law without integration cause charge density isn't constant, have to integrate to find charge first. So derp, I got it now.
 

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