[Gauss' Law] Hollow insulating sphere?

AI Thread Summary
To find the electric field at r = 17.2 cm using Gauss' Law for a hollow insulating sphere with a uniform charge density, the charge enclosed must be calculated based on the volume between the inner and outer radii. The correct volume is determined by subtracting the inner volume from the outer volume, specifically using V = (4/3)(π)(0.1170^3 - 0.0658^3). The charge density of 79.9 μC/m³ is then multiplied by this volume to find the total charge enclosed. The confusion arose from incorrectly considering different volumes for the calculations. The final understanding is that the entire charge is enclosed within the sphere of radius r.
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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