How Does Electric Field Behave Inside a Uniformly Charged Sphere?

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SUMMARY

The discussion focuses on the behavior of the electric field (E) inside a uniformly charged sphere, specifically under Gauss' law. It concludes that as one penetrates the sphere, the electric field decreases due to the diminishing enclosed charge, which is calculated using the formula q = charge density * (4/3)πr³. The analysis shows that while the electric field is influenced by the distance from the center (r² in the denominator), the decrease in charge (q) dominates, leading to a linear decrease in the electric field as the radius (r) decreases.

PREREQUISITES
  • Understanding of Gauss' law in electrostatics
  • Familiarity with electric field concepts
  • Knowledge of charge density and its calculations
  • Basic calculus for understanding the relationships between variables
NEXT STEPS
  • Study the implications of Gauss' law in different charge distributions
  • Explore the concept of electric fields in non-uniform charge distributions
  • Learn about the mathematical derivation of electric fields using integration techniques
  • Investigate the relationship between electric field strength and potential energy
USEFUL FOR

Students of physics, particularly those studying electrostatics, educators teaching electric field concepts, and anyone interested in the mathematical foundations of electric fields in charged objects.

theowne
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Under the Gauss' law section of the book.

Homework Statement


As you penetrate a uniform sphere of charge, E should decrease as less charge is inside the sphere, while E should increase because you are closer to the center of this charge. Which effect dominates?


Homework Equations


εΦ = q


The Attempt at a Solution



This is what I was thinking, maybe I'm completely misguided. q = charge density * 4/3 pi r^3 so q decreases by a factor of r^3. From gauss law' E = q / ε 4 pi r^2. So E decreases with q which decreases by r^3 while E increases with r^2 factor due to the denominator. So I would guess that q is the dominant factor there. Would it be a saisfactory answer if I just did [charge density * 4/3 pi r^2] / [ε 4 pi r^2] = (charge density * r) / (3 ε) = E so in the end as r decreases the E field will decrease.
 
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Looks great to me! E decreases linearly as r decreases.
 

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