Using Gauss' Law on a Solid Annular Sphere

[deathsink]

Homework Statement

Imagine a solid, annular sphere. At the center of the hollow is a point charge +Q. The inner radius of the sphere is r0, and the outer radius is R. Assume the charge density p = p0/r (for r0 < r < R). Calculate using the integral form of Gauss's Law the electric field in all three regions:

0< r < r0
r0< r < R
R < r < infinity

Homework Equations

[PLAIN]http://www.forkosh.dreamhost.com/mimetex.cgi?%5Coint_S%5Cmathbf%7BE%7D%5Ccdot%7Bd%7D%5Cmathbf%7BA%7D=%5Cfrac%7BQ%7D%7B%5Cepsilon_0%7D=%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D%5Cint_V%5Crho(%5Cmathbf%7Br%7D)%5C,dV

The Attempt at a Solution

for 0 < r < r0

E(4 Pi r^2) = 1/e0 (p0/r)(4/3 pi r^3)
=> E = (1/3e0)p0

But this can't be right for because all of the r's cancel out. And since I can't do this one, I know I can't do the rest.

Thank you.

 

[gneill]

For this first interval, where 0 < r < r0, the enclosed charge is just +Q, the charge at the center of the hollow.

 

[deathsink]

For this first interval, where 0 < r < r0, the enclosed charge is just +Q, the charge at the center of the hollow.

When you take the surface integral of E*dA, do you get E*(area of sphere) which you then need to divide through by?

I have not had Vector Calculus, so a lot of the notation is odd to me.

Edit: I got $$E=\frac{Q}{4\pi \epsilon _0r_0{}^2}$$ by dividing through by the area of a sphere.

 

[gneill]

Your E dot DA in the surface integral turns into E da, because the field is radially directed and so the dot product is just the product of the magnitudes of the vectors E and dA (which is normal to the surface over which you're integrating).

So, at radius r you end up with

E(4 π r²) = Q/ε₀

Or, rearranged,

E = Q/(4 π ε₀ r²)

Which should look familiar.

 

[paranormal]

I would first like to clarify that Gauss' Law is a fundamental principle in electromagnetism that relates the electric field to the distribution of electric charges. It states that the flux of the electric field through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space.

In this problem, we are given a solid, annular sphere with a point charge +Q at its center. The inner radius of the sphere is r0 and the outer radius is R. We are also given the charge density p = p0/r for r0 < r < R. The task is to calculate the electric field in all three regions: 0 < r < r0, r0 < r < R, and R < r < infinity.

Let us start with the region 0 < r < r0. In this region, the electric field is zero since there are no charges enclosed by the surface. This can be seen from Gauss' Law, where the total charge enclosed is 0 and therefore the flux through the surface must also be 0.

Moving on to the region r0 < r < R, we can use Gauss' Law to calculate the electric field. We will choose a Gaussian surface in the shape of a cylinder with the axis along the z-direction and the top and bottom surfaces at r = r and r = R, respectively. The side surface will be at a distance r from the z-axis. The symmetry of the problem allows us to assume that the electric field is constant on this surface and directed radially outward. The flux through this surface is given by:

Φ = E(2πrl) = Qenc/ε0

where Qenc is the enclosed charge and l is the length of the cylinder. Since the charge density is given by p = p0/r, the enclosed charge is:

Qenc = ∫p dV = 2πp0∫r dr = 2πp0(r^2 - r0^2)

Substituting this into Gauss' Law, we get:

E(2πrl) = (2πp0(r^2 - r0^2))/ε0

Solving for E, we get:

E = (p0(r^2 - r0^2))/(ε0l)

Since l = R - r0, we can write the electric field as:

E = (p0(r^