# Find the uniform charge density

DODGEVIPER13

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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DODGEVIPER13

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DODGEVIPER13
Oh and the question is 2.3 sorry should have mentioned this too

DODGEVIPER13
ok I can upload my work now

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Homework Helper
Gold Member
Hard to read your work, and looks strange. At any rate my answer was quite different and negative.

DODGEVIPER13
Yah I figured I was wrong any hints?

Homework Helper
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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.

DODGEVIPER13
double integral of E dot Ds = Q/epsilon0

Homework Helper
Gold Member
double integral of E dot Ds = Q/epsilon0

No. E does not appear. Just D, area and q.

DODGEVIPER13
Ok closed surface integral of E dot dA = Qfree

DODGEVIPER13
Whoops not E. I meant D

DODGEVIPER13
D dot dA=Qfree

Homework Helper
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D dot dA=Qfree

Excellent.

Now, how do you determine the total free charge inside the Gaussian surface at 3m?

DODGEVIPER13
Ok well should I guess I need to figure out limits of integration do the limits ave anything to do with (pi)r^2

Homework Helper
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You need to determine the total free charge inside the Gaussian surface at r = 3m. No integration required.

DODGEVIPER13
Hmm well what would D be here 2 because of 8-6 then 2x(pi)(3)^2

Homework Helper
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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!

DODGEVIPER13
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

Homework Helper
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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.

DODGEVIPER13
Ok so 8(pi)(1)^2-6(pi)(2)^2=-16(pi)

DODGEVIPER13
Am I good sorry not trying to rush you just wondering?

DODGEVIPER13
So ar that is

Homework Helper
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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.

DODGEVIPER13
(8E-9)(4)(pi)(1)^2-(6E-9)(4)(pi)(2)^2=-2.01E-7 C/m^2

Homework Helper
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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.

DODGEVIPER13
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

Homework Helper
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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!