Gauss Law

  • Thread starter Pruddy
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  • #1
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1. Homework Statement

-38.0 nC of charge is uniformly distributed throughout a spherical volume of radius 34.0 cm.
How much charge is contained in a region of radius 23.0 cm concentric with the charge distribution?


Homework Equations



Charge density = λ/area

The Attempt at a Solution



I don't know how to approach this problem. I dont know if it is right to use gaus's law.Please if anyone can give me directions or what topic in physics this problem comes from, I will gladly appreciate. Thanks
 

Answers and Replies

  • #2
tiny-tim
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Hi Pruddy! :smile:
-38.0 nC of charge is uniformly distributed throughout a spherical volume of radius 34.0 cm.
How much charge is contained in a region of radius 23.0 cm concentric with the charge distribution?

You're told that the distribution is uniform, so this is just geometry

"-38.0 kg of cheese is uniformly distributed throughout a spherical volume of radius 34.0 cm.
How much cheese is contained in a region of radius 23.0 cm concentric with the cheese distribution?" :biggrin:
 
  • #3
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:confused: Tiny-Tim,
So am I to used the Electric flux formula to solve this problem? But the question is looking for how much charge?
 
  • #4
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This question has nothing to do with the Gauss's Law or electric flux ( since it doesn't mention the electric field ).

As pointed out in post #2, this is a geometry problem.

Imagine starting with your sphere and removing a section with radius 23cm.
What is the relation between the removed part and the whole sphere?
 
  • #5
Doc Al
Mentor
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Charge density = λ/area
What you need is charge density ρ = Q/volume.

How does the total volume (radius 34 cm) compare to the volume of the region (radius 23 cm)?
 
  • #6
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Set up a Ratio, [itex]\frac{-38nc}{V_{1}}[/itex]=[itex]\frac{Q_{2}}{V_{2}}[/itex]
 
  • #7
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Hi, whynot314.
I used the formular and this are my calculations:

q2 = -38 x 10^(-9)/4*pi*r(0.34)^2 = q2/(4*pi*r(0.23)^2

= -38 x 10^(-9)*4*pi*r(0.23)^2 /(4*pi*r(0.34)^2
= -2.293 x 10^(-9)
I dont know if this is right...
 
  • #8
Dick
Science Advisor
Homework Helper
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Hi, whynot314.
I used the formular and this are my calculations:

q2 = -38 x 10^(-9)/4*pi*r(0.34)^2 = q2/(4*pi*r(0.23)^2

= -38 x 10^(-9)*4*pi*r(0.23)^2 /(4*pi*r(0.34)^2
= -2.293 x 10^(-9)
I dont know if this is right...

You should be using volumes, not surface areas. This isn't a Gauss' law problem. It's a charge density problem. The volume of a sphere of radius r is (4/3)pi*r^3. But even using area the numbers still don't come out the way you say they do.
 
Last edited:
  • #9
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Hi, whynot314,
Thanks a lot! You are the best.
 
  • #10
64
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Hi, whynot314,
I got it now.
 
  • #11
76
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I am studying this stuff right now to so this problem helped me as well.
 
  • #12
64
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wow. That's awesome!
 

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