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Gauss' Law: Spherical Symmetry

  • Thread starter MikeB2210
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  • #1
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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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Answers and Replies

  • #2
BvU
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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 ?
 
  • #3
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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?
 
  • #4
BvU
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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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