[Gauss' Law] Hollow insulating sphere?

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SUMMARY

The discussion focuses on applying Gauss' Law to determine the electric field outside a hollow insulating sphere with an inner radius of 6.58 cm and an outer radius of 11.7 cm, which has a uniform charge density of 79.9 μC/m³. The correct approach to find the "charge enclosed" involves calculating the volume of the charge distribution using the formula V = (4/3)π(b³ - a³), where 'b' is the outer radius and 'a' is the inner radius. The user confirms that the charge is uniformly distributed only within the hollow region, leading to the conclusion that the entire charge is enclosed within a sphere of radius r = 17.2 cm.

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  • Understanding of Gauss' Law and its application in electrostatics.
  • Familiarity with the concept of charge density and its calculation.
  • Knowledge of volume calculations for spherical objects.
  • Basic algebra and geometry skills for manipulating equations.
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  • Learn about electric field calculations for spherical charge distributions.
  • Explore the concept of electric flux and its relation to charge enclosed.
  • Investigate the implications of charge density variations in electrostatic problems.
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Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electrostatics and Gauss' Law applications.

GeorgeCostanz
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Homework Statement



A hollow insulating sphere with an inner radius of 6.58 cm and outer radius of 11.7 cm has a uniform charge density of 79.9 μC/m3 distributed throughout the volume between.

If we want to use Gauss' Law to find the electric field at r = 17.2 cm, what "charge enclosed" should we use?

Homework Equations



RHO = Q/V

a = .0658 m
b = .1170 m
r = .1720 m

therefore r > b > a

The Attempt at a Solution



i kno RHO = Q/V

i just don't kno what V to use

i've used v = 4/3(pi)(.1720^3)

i've used v = (4/3)(pi)(.1720^3 - .1170^3)

think I've used every combination of r

i kno I'm overlooking something incredibly simple
blah!
 
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The charge is only in the middle region and the entire charge is enclosed in a sphere of radius r.

ie,v = (4/3)(pi)(.1170^3 - .0658^3)
 
yes that makes perfect sense
and i could have swore i used those r's for v

thanks humanist
 

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