Need Help With Guass Laws and Amperes Law

In summary: The charge within a surface with r > b is just the charge at the center (+q) added to the charge on the conductor, which in this case is -2q.
  • #1
AdiV
15
0
Hi, I am really stuck with these problems, is it possible you guys can help me out, by guiding me along this problem, I wil write the entire thing out, but would appreciate it if you guys help me step by step so that I can understand it, since I am having a VERY hard time with these problems.

Homework Statement


A point charge (+q) is located at the origin. A spherical conductor of inner radius "a" and outer radius "b" surrounds this point charge with its center at the origin. A total charge of =2q lies on this spherical conductor.
{I put the diagram as an attachment}


a) Find the magnitude of the electric field in region I, at a distance r<a away from the origin. Label on the diagram some E field line within this region, if necessary. [Hint: Use Guass Laws]

b) Find the magnitude of the electric field in region II, at a distance r>1 (but <b), from the origin. Again label on the diagram some E fields if necessary.

c) Find the magnitude of the electric field in region III, at a distance r>b from teh origin. Again label on the diagram E field lines if necesary.


Consider 2 different designs for a printer cable. Design 1 has 4 cables each carrying a current I into the page. Design 2 has 4 cables all carrying current magnitude I but two have the current in one direction and two have in the opposite direction.

d) Find the magnetic field at a distance r. (in the plane which is perpendicular to the wires), away from cable 1 by using Ampere's Law.

e) Find the magnetic field a distance r away from cable 2 by using AMpere's LAw.

f) Which design is better, why?

Homework Equations



Guass Law;
[tex]\Sigma[/tex] Eperp * delta A = Qenclosed / [tex]\epsilon[/tex]0

Ampere's Law;
[tex]\Sigma[/tex] Bparallel * delta l = [tex]\mu[/tex]0 * Ienclosed

The Attempt at a Solution



Again, thank you for the help here.
 

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  • #2
Ok, I 've been working on it and well, I got to this point, but am not too sure.

a) sigma E perpendicular delta A = E * sigma delta A = E(4*pi*r^2) = Q / epsilon

So wouldn't the charge in section I be +q / epsilon?

c) sigma E perpendicular to delta A = E * sigma A = E * (4 pi r^2) = Q / epsilon

so E = 1 / [4 pi r^2] * Q / b

so, doesn't this tell us that the field outside the uniformly charged spherical shell is the same as if all the charge were concentrated at the center as a point charge??

So that would mean that it is -2q?
I am not too sure
 
  • #3
AdiV said:
Ok, I 've been working on it and well, I got to this point, but am not too sure.

a) sigma E perpendicular delta A = E * sigma delta A = E(4*pi*r^2) = Q / epsilon

So wouldn't the charge in section I be +q / epsilon?
The charge contained within that surface is given as +q. So solve for E.

c) sigma E perpendicular to delta A = E * sigma A = E * (4 pi r^2) = Q / epsilon

so E = 1 / [4 pi r^2] * Q / b

so, doesn't this tell us that the field outside the uniformly charged spherical shell is the same as if all the charge were concentrated at the center as a point charge??
That's correct. (The formula has a typo: You have b instead of epsilon.)

So that would mean that it is -2q?
What's the total charge within the Gaussian surface when r > b?
 
  • #4
The total charge is + q?

Can we go over this a bit more? I am really feeling lost right now.
 
  • #5
AdiV said:
The total charge is + q?
To find the total charge within a surface with r > b, just add the charge at the center (+q) to the charge on the conductor. (What's the charge on the conductor?)
 
  • #6
The charge on the conductor is -2q, so +q + -2q will be -q so that is the charge in region I?
 
  • #7
AdiV said:
The charge on the conductor is -2q, so +q + -2q will be -q so that is the charge in region I?
No, that's the charge within a sphere of radius r > b. Region I is within r < a; the only charge in that region is the center charge of +q.

Note: The regions are for the purpose of defining the field. For finding the charge, use a Gaussian sphere of the appropriate radius, counting all the charge within that radius.
 
  • #8
Ohh, ok, I think I understand now, so for region II the charge would be -2q since that is the charge of the conductor
 
  • #9
AdiV said:
Ohh, ok, I think I understand now, so for region II the charge would be -2q since that is the charge of the conductor
Careful here: What's the field in region II? (Hint: What's the field within a conductor?)

While the total charge on the conductor is -2q, where is that charge located? What's the charge on its inner surface? Its outer surface?
 
  • #10
The E field in a conductor is zero right?
 
  • #11
AdiV said:
The E field in a conductor is zero right?
Exactly. So part (b) is easy.
 
  • #12
Wow, thank you so much!
 
  • #13
AdiV said:
Wow, thank you so much!
Just to prove to yourself that you fully understand it, try to answer my other questions in post #9. (Use Gauss's law.)
 

1. What are Gauss' Laws and Ampere's Law?

Gauss' Laws and Ampere's Law are fundamental laws in electromagnetism that describe the behavior of electric and magnetic fields. Gauss' Laws state that the electric flux through a closed surface is proportional to the charge enclosed by that surface, while Ampere's Law states that the magnetic field around a closed loop is proportional to the electric current passing through that loop.

2. How are Gauss' Laws and Ampere's Law related?

Gauss' Laws and Ampere's Law are closely related as they both describe the fundamental behavior of electric and magnetic fields. They are complementary laws and are often used together to understand and analyze electromagnetic phenomena.

3. How can I apply Gauss' Laws and Ampere's Law in practical situations?

Gauss' Laws and Ampere's Law can be applied in various practical situations, such as in the design of electrical circuits, electromagnetic devices, and in understanding the behavior of electromagnetic waves. They are also used in many engineering and scientific fields, including telecommunications, power systems, and medical imaging.

4. Are there any limitations to Gauss' Laws and Ampere's Law?

While Gauss' Laws and Ampere's Law are powerful tools for understanding electromagnetic phenomena, they have some limitations. For example, they are only applicable to static electric and magnetic fields, and cannot fully describe the behavior of time-varying fields. Additionally, they do not take into account quantum effects and relativistic effects.

5. Can Gauss' Laws and Ampere's Law be derived from other fundamental laws?

Yes, Gauss' Laws and Ampere's Law can be derived from other fundamental laws, such as Coulomb's Law and the Biot-Savart Law. They are also closely related to Maxwell's Equations, which are a set of four equations that describe the behavior of electric and magnetic fields in a more comprehensive manner.

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