Find the uniform charge density

  • #1
DODGEVIPER13
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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 cant 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 dont know what to do?
 
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Answers and Replies

  • #2
DODGEVIPER13
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opps forgot to upload
 

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  • #3
DODGEVIPER13
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Oh and the question is 2.3 sorry should have mentioned this too
 
  • #4
DODGEVIPER13
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ok I can upload my work now
 

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  • #5
rude man
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Hard to read your work, and looks strange. At any rate my answer was quite different and negative.
 
  • #6
DODGEVIPER13
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Yah I figured I was wrong any hints?
 
  • #7
rude man
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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
DODGEVIPER13
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double integral of E dot Ds = Q/epsilon0
 
  • #10
DODGEVIPER13
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Ok closed surface integral of E dot dA = Qfree
 
  • #11
DODGEVIPER13
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Whoops not E. I meant D
 
  • #12
DODGEVIPER13
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D dot dA=Qfree
 
  • #14
DODGEVIPER13
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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
rude man
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You need to determine the total free charge inside the Gaussian surface at r = 3m. No integration required.
 
  • #16
DODGEVIPER13
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Hmm well what would D be here 2 because of 8-6 then 2x(pi)(3)^2
 
  • #17
rude man
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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
DODGEVIPER13
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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
rude man
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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
DODGEVIPER13
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Ok so 8(pi)(1)^2-6(pi)(2)^2=-16(pi)
 
  • #21
DODGEVIPER13
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Am I good sorry not trying to rush you just wondering?
 
  • #22
DODGEVIPER13
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So ar that is
 
  • #23
rude man
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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
DODGEVIPER13
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(8E-9)(4)(pi)(1)^2-(6E-9)(4)(pi)(2)^2=-2.01E-7 C/m^2
 
  • #25
rude man
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(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
DODGEVIPER13
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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
rude man
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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!
 

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