Find the uniform charge density

In summary, the student attempted to solve a homework problem involving the determination of the total free charge inside a Gaussian surface at a given radius, but got lost. After some basic algebra, he figured out that the charge would be -1.777778 nC/m^2.
  • #1
672
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Homework Statement


Upload Problem 2.3 and I uploaded the question in a reply below

Homework Equations


D/2ε
E=KQ/R^2
E=E1+E2+E3

The Attempt at a Solution


I would upload my work but I am in a location where I can't right now but I can type what I did hopefully it will be clear enough. When r is outside of the sphere E=KQ/r^2 and when r is inside the sphere E=KQr/a^3. Well when r=1 it is inside and when r=2 it is inside but however at r=3 it is outside I assume then to solve the problem I need to set D/2ε=KQ/r^2 or some combination of this to get it correct, however because the equation adds the "a" term. Which is shown in the problem but is not given I don't know what to do?
 
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  • #2
opps forgot to upload
 

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  • #3
Oh and the question is 2.3 sorry should have mentioned this too
 
  • #4
ok I can upload my work now
 

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  • #5
Hard to read your work, and looks strange. At any rate my answer was quite different and negative.
 
  • #6
Yah I figured I was wrong any hints?
 
  • #7
DODGEVIPER13 said:
Yah I figured I was wrong any hints?

Sure thing.

Start with the basic formula relating the D flux across a closed surface and the free charge within the volume circumscribed by that surface.
 
  • #8
double integral of E dot Ds = Q/epsilon0
 
  • #9
DODGEVIPER13 said:
double integral of E dot Ds = Q/epsilon0

No. E does not appear. Just D, area and q.
 
  • #10
Ok closed surface integral of E dot dA = Qfree
 
  • #11
Whoops not E. I meant D
 
  • #13
DODGEVIPER13 said:
D dot dA=Qfree

Excellent.

Now, how do you determine the total free charge inside the Gaussian surface at 3m?
 
  • #14
Ok well should I guess I need to figure out limits of integration do the limits ave anything to do with (pi)r^2
 
  • #15
You need to determine the total free charge inside the Gaussian surface at r = 3m. No integration required.
 
  • #16
Hmm well what would D be here 2 because of 8-6 then 2x(pi)(3)^2
 
  • #17
The 8 and -6 are surface charge densities, not charges. You need to compute the total net charge as seen by the Gaussian spherical shell at r = 3m.

Don't think about D until you have the net free charge. One step at a time!
 
  • #18
Well how about this 8/(2(epsilon))=Kq1/(1)^2 and 6/(2(epsilon))=Kq2/(2)^2 where q1=16(pi) and q2= -48(pi) and then q3=q1+q2 so -32(pi) for Qfree? am I close I mean U figure I ave to use the surface carge densities in dome way to find the total charge
 
  • #19
DODGEVIPER13 said:
Well how about this 8/(2(epsilon))=Kq1/(1)^2 and 6/(2(epsilon))=Kq2/(2)^2 where q1=16(pi) and q2= -48(pi) and then q3=q1+q2 so -32(pi) for Qfree? am I close I mean U figure I ave to use the surface carge densities in dome way to find the total charge

Yes, but not that way!

Charge = charge density times area. k and epsilon have nothing to do with this problem.
 
  • #20
Ok so 8(pi)(1)^2-6(pi)(2)^2=-16(pi)
 
  • #21
Am I good sorry not trying to rush you just wondering?
 
  • #23
DODGEVIPER13 said:
Ok so 8(pi)(1)^2-6(pi)(2)^2=-16(pi)

No. But don't be discouraged, you're getting warm.

What is the area of a spherical surface? It's not pi(r^2).

And the densities are 8 and -6 nC/m^2, not 8 and -6 C/m^2.C.
 
  • #24
(8E-9)(4)(pi)(1)^2-(6E-9)(4)(pi)(2)^2=-2.01E-7 C/m^2
 
  • #25
DODGEVIPER13 said:
(8E-9)(4)(pi)(1)^2-(6E-9)(4)(pi)(2)^2=-2.01E-7 C/m^2

The right-hand side is charge, not charge density. I am not checking your arithmetic but otherwise the rhs is OK if you fix the units.

So, we now tackle the other side of Gauss' law as applied to the spherical surface with radius = 3m.
 
  • #26
ok so with fixed units it should be (8)(4)(pi)(1)^2-(6)(4)(pi)(2)^2=-64(pi) nC and so the otherside would be D(4)(pi)(3)^2 which is 36(pi)D=-64(pi) so D=-64/36=-1.777778 nC/m^2
 
  • #27
DODGEVIPER13 said:
ok so with fixed units it should be (8)(4)(pi)(1)^2-(6)(4)(pi)(2)^2=-64(pi) nC and so the otherside would be D(4)(pi)(3)^2 which is 36(pi)D=-64(pi) so D=-64/36=-1.777778 nC/m^2

Very good! Unless ther is an arithmetic error, this is correct - even the units!
 

What is uniform charge density?

Uniform charge density refers to the distribution of electric charge that is constant throughout a given region.

How do you find the uniform charge density?

To find the uniform charge density, you divide the total charge by the volume of the region. This gives you the amount of charge per unit volume, which is the uniform charge density.

Why is finding the uniform charge density important?

Finding the uniform charge density is important for understanding and analyzing electric fields and the behavior of charged particles in a given region. It also allows for the calculation of electric potential and other important parameters.

What units are used for uniform charge density?

The SI unit for uniform charge density is coulombs per cubic meter (C/m³). Other commonly used units include coulombs per liter (C/L) and coulombs per cubic centimeter (C/cm³).

What are some real-world applications of finding uniform charge density?

Finding uniform charge density has practical applications in fields such as electrical engineering, physics, and chemistry. It is used in the design of electronic devices, the study of electrostatics and magnetostatics, and the analysis of chemical reactions involving charged particles.

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