Need Help With Guass Laws and Amperes Law

[AdiV]

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;
\Sigma Eperp * delta A = Qenclosed / \epsilon0

Ampere's Law;
\Sigma Bparallel * delta l = \mu0 * Ienclosed

The Attempt at a Solution

Again, thank you for the help here.

 

[AdiV]

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

 

[Doc Al]

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?

 

[AdiV]

The total charge is + q?

Can we go over this a bit more? I am really feeling lost right now.

 

[Doc Al]

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?)

 

[AdiV]

The charge on the conductor is -2q, so +q + -2q will be -q so that is the charge in region I?

 

[Doc Al]

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.

 

[AdiV]

Ohh, ok, I think I understand now, so for region II the charge would be -2q since that is the charge of the conductor

 

[Doc Al]

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?

 

[AdiV]

The E field in a conductor is zero right?

 

[Doc Al]

The E field in a conductor is zero right?

Exactly. So part (b) is easy.

 

[AdiV]

Wow, thank you so much!

 

[Doc Al]

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.)