# Coaxial Cylindrical Conductors

1. Feb 5, 2008

### heartofaragorn

1. The problem statement, all variables and given/known data
Two coaxial cylindrical conductors are shown. The inner cylinder has radius a = 2 cm, length 10 m, and carries a total charge of Q inner = +8nC. The outer cylinder has an inner radius b = 6 cm, outer radius c= 7 cm, length 10 m, and carries a total charge of Q outer = -16nC. What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r= 4 cm from the origin and makes an angle of 30 degrees with the x-axis?

2. Relevant equations
Surface charge density = 2pi * radius a * length
Linear charge density = 2pi * radius a * surface charge density
Electric field = 2 * k * linear charge denisty / r

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 5, 2008

### Mindscrape

So what direction does the electric field, from Gauss's law, travel in? I'm guessing this is where are getting tripped up.

Also, do you have a picture? I can't tell why they even give you an outer cylinder, something you mentioned also, maybe the problem creator is just trying to throw you off. Could you repost your work but in terms of symbols rather than numbers?

3. Feb 6, 2008

### heartofaragorn

I'll try attaching the image, but as for the symbols, I don't have anything that will display them. My computer is a little too old to have packages like Microsoft Equation or anything like that. If it's really a problem, I'll try seeing if I can download something. Also, I'm really not sure what direction the electric field travels in!

#### Attached Files:

• ###### Cylinder.gif
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Last edited: Feb 6, 2008
4. Feb 6, 2008

### Staff: Mentor

Your Gaussian cylinder at (radius r, say) totally encloses the inner conductor. What's Gauss's law? Set up the equation and solve for E. (No need to compute surface charge!)

The charge on the outer conductor is irrelevant. The field at r = 4 cm will be radially outward.

You can just type in your equations without using symbols, or you can use Latex. (See Introducing LaTeX Math Typesetting .)

5. Feb 6, 2008

### heartofaragorn

The interactive help that comes with the problem stated that I had to calculate the surface charge in order to calculate the linear charge. Also, do I have the right formula for E? And if I don't need the surface charge, from where do I get the linear charge?

6. Feb 6, 2008

### Staff: Mentor

That makes no sense to me. Using the data given, you can calculate the linear charge density directly.
Yes. (Use Gauss's law to verify it, if you're unsure.)
You are given the total charge and the length. (Assume it's uniformly distributed.)

7. Feb 6, 2008

### heartofaragorn

YEA! It worked!!!!!! Thanks so much!