Solving Gauss's Law Problem for Electric Field

In summary, the electric field in the regions: r_1 < a, a < r_2 < b, b < r_3 < c, and r_4 > c is -2Q/4π.
  • #1
PhysicsinCalifornia
58
0

Homework Statement


A uniformly charged ball of radius a and a total charge -Q is at the center of a hollow metal shell with inner radius b and outer radius c. The hollow sphere has a net charge +2Q. Find the magnitude of the electric field in the regions: [tex]r_1 < a[/tex],[tex] a < r_2 < b[/tex],[tex] b < r_3 < c[/tex], and [tex]r_4 > c.[/tex]

Homework Equations



[tex]V = \frac{4}{3} \pi R^3[/tex]
[tex]S = 4 \pi R^2[/tex]
[tex]\oint E(x)dA = \frac{q_{in}}{\epsilon_o}[/tex]

The Attempt at a Solution



For E(r1 < a):
[tex]\rho = \frac{Q_{tot}}{\epsilon_o}[/tex]
[tex]Q_{in,tot} = \rho*\frac{4}{3} \pi r_1^3[/tex]
[tex]\oint_0^rE(x)dA = \frac{q_{in}}{\epsilon_o}[/tex]

[tex]E(r_1) = \frac{\rho\frac{4}{3} \pi r_1^3}{\epsilon*4 \pi r_1^2}[/tex]

[tex]E(r_1) = \frac{\rho*r_1}{3\epsilon_o}[/tex]

This is actually where I am stuck, I got everything else. Am I supposed to get rid of that volume charge density, [tex]\rho[/tex]?
 
Last edited:
Physics news on Phys.org
  • #2
Yes; in textbook problems, they expect you to express the answer in terms of the variables given in the statement of the problem.

The density is uniform and you know the total charge in the little ball. So just divide that by its volume to find its density.
 
  • #3
The following may help.

Regards,

Nacer.

http://islam.moved.in/tmp/c.jpg
 
Last edited by a moderator:
  • #4
There is actually a second and third part to this question:

b) Find potentials at points in the regions: [tex]r_1 < a[/tex], [tex]a < r_2 < b[/tex], [tex]b < r_3 < c[/tex], and [tex] r_4 >c[/tex]

For r1 < a,
I use the formula

[tex]V_o - V_{r_1} = \int_0^r E(r_1)dr[/tex]

[tex]E(r_1) = \frac{\rho*r_1}{3\epsilon_o}[/tex]

[tex]V_o - V_{r_1} = \frac{\rho}{3\epsilon_o} \int_0^r r dr[/tex]

[tex]V_o - V_{r_1} = \frac{\rho*r_1^2}{6\epsilon_o}[/tex]

Solving the same way, I got::

[tex]V_o - V_{r_2} = \frac{Q}{4 \pi \epsilon_o r_2}[/tex]

[tex]V_o - V_{r_3} = 0[/tex]

[tex]V_o - V_{r_4} = \frac{-Q}{4 \pi \epsilon_o r_4}[/tex]

Did anyone get the same answer as I did?

Also, I am supposed to sketch the graphs of how Ex depends on a distance(r) from the center of the sphere, and how V depends on a distance(r) from the center of the sphere.

I drew my graphs, but I don't know how to post it on. I don't have a scanner handy, so if anyone can help, please let me know what your graphs looks like.
Thank you very much, you are all so helpful.
 

1. What is Gauss's Law and how does it relate to electric fields?

Gauss's Law is a fundamental law in electromagnetism that relates the electric field at a point to the charge enclosed by a closed surface surrounding that point. In simple terms, it states that the electric flux through a closed surface is proportional to the amount of charge enclosed by that surface.

2. How do you apply Gauss's Law to solve problems involving electric fields?

To solve Gauss's Law problems for electric fields, you need to follow a few steps. First, identify the symmetry of the problem and choose an appropriate Gaussian surface. Then, determine the charge enclosed by the surface. Finally, use the formula for Gauss's Law to calculate the electric field at the point of interest.

3. Can you explain the concept of electric flux and its role in Gauss's Law?

Electric flux is a measure of the amount of electric field passing through a given surface. In Gauss's Law, it is used to relate the electric field at a point to the charge enclosed by a closed surface surrounding that point. Essentially, it allows us to calculate the electric field at a point without having to consider the individual contributions from all the charges in the system.

4. What types of problems can be solved using Gauss's Law for electric fields?

Gauss's Law can be applied to a wide range of problems involving electric fields, as long as there is some form of symmetry present. This can include problems with point charges, line charges, plane charges, and more complex charge distributions.

5. Are there any limitations to using Gauss's Law for solving electric field problems?

While Gauss's Law is a powerful tool for solving electric field problems, it does have its limitations. It can only be applied in situations where there is some form of symmetry present, and it does not take into account the effects of changing magnetic fields. Additionally, it can only be used to calculate the electric field at a point outside the charge distribution.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
485
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
814
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
4
Views
804
Replies
2
Views
699
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
23
Views
338
  • Advanced Physics Homework Help
Replies
13
Views
2K
Back
Top