What is the Charge on a Uniformly Charged Sphere Based on Gauss' Law?

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SUMMARY

The charge on a uniformly charged sphere can be determined using Gauss' Law, which states that the net electric flux through a closed surface is proportional to the enclosed charge. For a sphere with a uniform charge distribution, the electric field inside the sphere increases linearly with radius, while outside the sphere, it decreases with the square of the distance from the center. The relevant equations include the net flux equation and Gauss' Law, which relates the electric field to the enclosed charge. The charge can be calculated by integrating the electric field over the surface area of the sphere.

PREREQUISITES
  • Understanding of Gauss' Law and its application in electrostatics
  • Familiarity with electric field concepts and calculations
  • Knowledge of integration techniques in physics
  • Basic understanding of spherical coordinates and geometry
NEXT STEPS
  • Study the derivation of Gauss' Law in electrostatics
  • Learn how to calculate electric fields for different charge distributions
  • Explore the concept of electric flux and its applications
  • Investigate the relationship between electric field strength and distance from charged objects
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Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and Gauss' Law applications.

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Homework Statement


Figure 23.52 gives the magnitude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume. The scale of the vertical axis is set by Es = 5.0 x 10e7 N/C. What is the charge on the sphere?

Homework Equations


Net Flux = ∫E . dA
(e0)(Net Flux) = q(enclosed) (Gauss' Law)

The Attempt at a Solution

 

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Hello Mike, welcome to PF :)

Relevant eqns are OK. PF rules dictate that you post your working, so fill in the givens in the eqns and see how far you can come. Good chance you won't even need us any more ! Or is there something specific you want guidance for ?
 
I solved the problem but just have a question about the theory! Based on the figure given how do I know what part of the sphere I am at?
 
You can put Gauss theorem to work to answer that: enclosed q starts at zero for r=0 and increases as r3. Area increases as r2 so E increases as r1. When it stops increasing, you nust be at the surface of the sphere. From then on q doesn't increase, but A does and hence E goes like r-2

Good chance you won't even need us any more !
Pleased that my hunch was right :)
 

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