Calculating Charge in a Concentric Spherical Region

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The discussion revolves around calculating the charge contained in a concentric spherical region with a radius of 23.0 cm, given a uniformly distributed charge of -38.0 nC in a larger sphere of radius 34.0 cm. Participants clarify that this problem is primarily geometric and involves charge density rather than Gauss's Law. The correct approach is to use the volume ratio of the two spheres to determine the charge in the smaller sphere. The volume of a sphere is calculated using the formula (4/3)πr^3, and the charge density is defined as charge per unit volume. Ultimately, the conversation emphasizes understanding the relationship between the volumes and the uniform charge distribution.
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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 don't 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
 
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Hi Pruddy! :smile:
Pruddy said:
-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:
 
: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?
 
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?
 
Pruddy said:
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)?
 
Set up a Ratio, \frac{-38nc}{V_{1}}=\frac{Q_{2}}{V_{2}}
 
Hi, whynot314.
I used the formula 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 don't know if this is right...
 
Pruddy said:
Hi, whynot314.
I used the formula 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 don't 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:
Hi, whynot314,
Thanks a lot! You are the best.
 
  • #10
Hi, whynot314,
I got it now.
 
  • #11
I am studying this stuff right now to so this problem helped me as well.
 
  • #12
wow. That's awesome!
 
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