Solving Gauss Law Problem 1: Electric Field of a Charged Copper Shell and Wire

In summary: However, you then plugged in the value for lambda to an equation which said that the electric field at the surface of the wire is 10608.2/3.525E-5. This is incorrect - the electric field at the surface of the wire is 10608.2/3.525E-5 * 0.705 or 10608.2/3.525E-5 * 3.525 or 30094.1844.
  • #1
purduegirl
74
0
1.

Homework Statement
A thin, cylindrical copper shell of diameter ds = 6 cm has along its axis a thin metal wire of diameter dw = 7.05 × 10-3 cm. (as shown in the diagram below.) The wire and the shell carry equal and opposite charges of 5.90 × 10-9C/cm, distributed uniformly. What is the magnitude of the electric field at a point r = 3.30 mm, where r is measured radially from the center of the inner wire? What is the magnitude of the electric field at the surface of the wire?

Homework Equations


I know this is refers to Gauss law.


The Attempt at a Solution


For the first part, I know that I need to find the Electric Field E = the charge/ 2*pie*epilson zero, the radius. My prof said to take this answer into the diameter, but I'm not sure what he means by that. The second part I have no idea how to do.
 
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  • #2
I think you are trying to apply a result derived from Gauss' law without really understanding Gauss' law. Take a cylinder of radius r and length L centered on the wire with r between the radius of the wire and the shell. What's the charge contained in the cylinder? What's the electric flux through the cylinder (considering that the E field is constant through the outer surface of cylinder and the surface area of that outer surface is 2*pi*r)? What does Gauss' law say the relation between the two is?
 
  • #3
To give you more insight into this kind of problem, I recommend a book by W.H. Hayt called Engineering Electromagnetics, a very readable text. I bet you can find a copy over in Potter EC library. (If you are curious about this author. There is a wall display, dedicated to him at the Mem'l Union).
 
  • #4
What is the magnitude of the electric field at the surface of the wire?

I know that the E field = 2*k*lambda/r

I am given lambda which I converted to m which I get 5.90E-7 C/m.
I am given the diameter of the wire = 7.05E-3 cm or 0.705 m. The radius will be 0.3525 m. I found that the E field using this formula is 3.01E4 N/C. This is wrong. Is this even the right equation?
 
  • #5
purduegirl said:
I am given the diameter of the wire = 7.05E-3 cm or 0.705 m.
Careful here! E-3 cm = E-5 m.
 
  • #6
I don't see what equation you are using, but the unit conversions are not going so well. 7.05E-3cm is NOT 0.705m.
 
  • #7
the answer I got was 10608.2/3.525E-5 = .030094 N/C, still incorrect
 
  • #8
purduegirl said:
the answer I got was 10608.2/3.525E-5 = .030094 N/C, still incorrect

That's not good. What numbers are you plugging into what equation?
 
  • #9
Dick said:
I don't see what equation you are using, but the unit conversions are not going so well. 7.05E-3cm is NOT 0.705m.

I know that the E field = 2*k*lambda/r
 
  • #10
2(8.99e9)(5.90e-11)/(3.525e-5) = 30094.1844
 
  • #11
You said before lambda=5.90E-7 C/m. That conversion was correct.
 

1. What is Gauss Law and why is it important in solving electric field problems?

Gauss Law is a fundamental law in electromagnetism that relates the distribution of electric charges to the resulting electric field. It is important in solving electric field problems because it provides a mathematical framework for understanding and predicting the behavior of electric fields.

2. What is the formula for Gauss Law and how is it applied in solving electric field problems?

The formula for Gauss Law is ∮E⋅dA = Qenc / ε0, where ∮E⋅dA represents the electric flux through a closed surface, Qenc is the total charge enclosed by that surface, and ε0 is the permittivity of free space. This formula is applied by using it to calculate the electric field at a given point due to a known distribution of charges.

3. What is the setup for Gauss Law Problem 1 and how do we approach solving it?

Gauss Law Problem 1 involves finding the electric field due to a charged copper shell and wire. The setup includes a spherical shell made of copper with a charge Q on its surface, and a straight copper wire passing through the center of the shell with a charge -Q. To solve this problem, we use the formula for Gauss Law and apply it to a closed surface enclosing either the shell or the wire, depending on which electric field component we are trying to find.

4. What are the key steps in solving Gauss Law Problem 1?

The key steps in solving Gauss Law Problem 1 are:

  1. Choose a closed surface to apply Gauss Law to, based on which electric field component we are trying to find.
  2. Calculate the electric flux through that surface using the formula ∮E⋅dA.
  3. Determine the total charge enclosed by the surface, Qenc.
  4. Substitute the values into the formula ∮E⋅dA = Qenc / ε0 and solve for the electric field component.

5. What are some common mistakes to avoid when solving Gauss Law Problem 1?

Some common mistakes to avoid when solving Gauss Law Problem 1 include:

  • Using the wrong closed surface or not enclosing the correct charges in the chosen surface.
  • Forgetting to take into account the direction of the electric field and the orientation of the surface.
  • Not considering the symmetry of the problem and using the wrong formula or approach.
  • Not converting units correctly when necessary.

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