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

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Homework Help Overview

The discussion revolves around determining the charge on a uniformly charged sphere using Gauss' Law, with a focus on understanding the electric field behavior both inside and outside the sphere.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Gauss' Law and the relationship between the electric field and the radius within the sphere. Questions arise regarding how to identify the specific region of the sphere being analyzed based on the provided figure.

Discussion Status

Some participants have provided insights into the theoretical aspects of the problem, particularly regarding the behavior of electric field strength in relation to the radius. There is an ongoing exploration of the implications of these relationships without a clear consensus on the specific charge value.

Contextual Notes

Participants are encouraged to provide their working details as per forum rules, and there is an emphasis on understanding the theoretical foundations rather than just arriving at a numerical solution.

MikeB2210
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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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