Advanced Gauss's Law Question

The charge enclosed would be the sum of charges from both the charge densities i.e \rho and \sigma. Then plug in the respective expression for the electric field inside a sphere in terms of the charge density, and try to equate that to the expression for the electric field outside the sphere. That should give you Q_{total}.In summary, to find the values of \rho and \sigma in terms of R, r, and Q_{volume}, we use Gauss's Law to set up Gaussian surfaces inside and outside the sphere of charge. By equating the electric field expressions inside and outside the sphere, we can solve for Q_{total} and then use it to find the values of \rho and \sigma.
  • #1
blackhawk97
2
0
A solid sphere of radius [tex]R[/tex] has a non-uniform volume charge density [tex]\rho(r)[/tex] and a constant surface charge density [tex]\sigma[/tex]. If the field inside the sphere is uniform and radially atuned, and the field a distance [tex]2R[/tex] away from the center is zero, find [tex]\rho[/tex] and [tex]\sigma[/tex] in terms of [tex]R[/tex], [tex]r[/tex] (distance from the center of the sphere), and [tex]Q_\text{volume}[/tex] (the charge associated with [tex]\rho[/tex], but not with [tex]\sigma[/tex]).

Homework Equations


Gauss's Law

The Attempt at a Solution


I'm not sure how to proceed, but I think the solution should begin by find the total charge on the sphere (ie., adding the integral of the charge calculable from the surface charge density with the integral of the charge calculable from the volume charge density). Am I on the right track?
 
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  • #2
So, inside the sphere, what does Gauss's law say? Outside the sphere? Try drawing Gaussian surfaces which are spheres inside and outside the sphere of charge.
 
  • #3
I know how to do the first part, i think... (finding [tex]\sigma[/tex]). Basically you just use Gauss's Law

[tex]\oint{E \cdot dA}=\frac{Q_\text{enclosed}}{\epsilon_0}[/tex]

... except you set the [tex]r[/tex] in that equation equal to [tex]2R[/tex], so you can ultimately set the expression for the [tex]E[/tex]-field equal to zero and... yeah.

But the part about [tex]\rho[/tex] still has me stumped. Can anyone offer a bit more help?

Also, correction to the problem: the field inside the sphere is not radially atuned, it is directed radially outwards.
 
  • #4
So can you express [tex]\sigma[/tex] in terms of [tex]Q_{total}[/tex]?

For the [tex]\rho[/tex] part you have to take Gauss surfaces inside the sphere as Matterwave said.
 
  • #5


Yes, you are on the right track. To find \rho and \sigma in terms of R, r, and Q_\text{volume}, we can use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε_0).

In this problem, we have a closed surface in the form of the sphere with radius 2R, and we know that the electric field is zero at a distance of 2R from the center. This means that the electric flux through the surface is also zero. We can then set up the following equation:

ϕ_E = Q_\text{enc}/ε_0 = 0

Since the electric flux is zero, this means that the enclosed charge (Q_\text{enc}) must also be zero. However, we know that there is a volume charge density \rho(r) and a surface charge density \sigma present within the sphere. So, we can set up the following equations:

Q_\text{enc} = Q_\text{volume} + Q_\text{surface}

Since we are given that the electric field inside the sphere is uniform and radially oriented, we can use Gauss's Law to find the electric field at any point within the sphere. This gives us the following equation:

E = Q_\text{enc}/(4πε_0r^2)

We can then substitute this into our previous equation and solve for Q_\text{volume}:

0 = Q_\text{volume} + Q_\text{surface} - (4πε_0r^2)E

Q_\text{volume} = (4πε_0r^2)E - Q_\text{surface}

Now, we know that the electric field at a distance of 2R from the center is zero, so we can substitute this in for r and solve for Q_\text{volume}:

Q_\text{volume} = (4πε_0(2R)^2)E - Q_\text{surface}

Q_\text{volume} = 16πε_0R^2E - Q_\text{surface}

We also know that the charge associated with \rho is equal to the volume charge density multiplied by the volume of the sphere, so we can write the following equation:

Q_\text{volume} = \rho(
 

1. What is Advanced Gauss's Law?

Advanced Gauss's Law is a mathematical principle that relates the electric field to the distribution of electric charges in a given space. It is an extension of the more basic Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of the medium.

2. How is Advanced Gauss's Law different from Gauss's Law?

Advanced Gauss's Law takes into account not only the enclosed charge, but also the current density within the closed surface. It also includes the displacement current, which is a time-varying electric field, and is often used in the study of electromagnetic waves.

3. What are some real-world applications of Advanced Gauss's Law?

Advanced Gauss's Law is used in many fields of science and engineering, including electromagnetism, electronics, and telecommunications. It is also an important principle in the study of optics and the behavior of light.

4. How is Advanced Gauss's Law used in the study of electromagnetic waves?

Advanced Gauss's Law is used to derive Maxwell's equations, which are a set of fundamental equations that describe the behavior of electric and magnetic fields in space. These equations are crucial in understanding the generation, propagation, and interaction of electromagnetic waves.

5. Are there any limitations to Advanced Gauss's Law?

Advanced Gauss's Law is a powerful tool in the study of electromagnetism, but it does have some limitations. It assumes a linear, isotropic, and homogeneous medium, which may not always be the case in real-world situations. It also does not take into account any quantum effects, which may be significant at very small scales.

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