# Find the uniform charge density

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?

Last edited:

Related Engineering and Comp Sci Homework Help News on Phys.org

#### Attachments

• 115.9 KB Views: 112
Oh and the question is 2.3 sorry should have mentioned this too

ok I can upload my work now

#### Attachments

• 7.2 KB Views: 324
rude man
Homework Helper
Gold Member
Hard to read your work, and looks strange. At any rate my answer was quite different and negative.

Yah I figured I was wrong any hints?

rude man
Homework Helper
Gold Member
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.

double integral of E dot Ds = Q/epsilon0

rude man
Homework Helper
Gold Member
double integral of E dot Ds = Q/epsilon0
No. E does not appear. Just D, area and q.

Ok closed surface integral of E dot dA = Qfree

Whoops not E. I meant D

D dot dA=Qfree

rude man
Homework Helper
Gold Member
D dot dA=Qfree
Excellent.

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

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

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

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

rude man
Homework Helper
Gold Member
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!

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

rude man
Homework Helper
Gold Member
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.

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

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

So ar that is

rude man
Homework Helper
Gold Member
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.

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

rude man
Homework Helper
Gold Member
(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.