# Gauss's Law and a conductor

1. Jan 30, 2010

### LonghornDude8

1. The problem statement, all variables and given/known data
The figure shows a portion of an infinitely long, concentric cable in cross section. The inner conductor carries a charge of 5.1 nC/m and the outer conductor is uncharged.
What is the electric field 1.9 cm from the central axis? The permittivity of free space is 8.85419e-12 C2/Nm2.
The picture:
inner cylinder has a diameter of 3.6 cm.
Outer hollowish cylinder: diameter1 = 8.2 cm; diameter2 = 10.7 cm

2. Relevant equations
$$\int EdA = Q_{ins}/\epsilon_{0}$$
Area of cylinder: $$A=2\pi r^{2}l$$

3. The attempt at a solution
Doing a little bit of algebra, I got the formula: $$\lambda/2\pi r\epsilon_{0}$$ and got the answer 4824.9 N/C.
I googled part of the question and saw that the way I solved it was right (my equation) and using my equation, I got their answer, but when I do the problem with the numbers I'm given, I keep getting it wrong.

Last edited: Jan 30, 2010
2. Jan 31, 2010

### gabbagabbahey

In what region?

3. Jan 31, 2010

### LonghornDude8

Using a cylinder as the Gaussian surface, you get:
$$E*2\pi rl = Q/\epsilon_{0}$$
and $$Q/l=\lambda$$
therefore, $$Q=l\lambda$$
substitute and divide both sides by $$2\pi rl$$:
$$E=l\lambda/2\pi rl\epsilon_0$$
cancel out the ls:
$$E=\lambda/2\pi r\epsilon_0$$

4. Jan 31, 2010

### jasper10

$$\mu$$ $$\sigma$$

5. Jan 31, 2010

### jasper10

oh sorry about that :S my mistake

6. Jan 31, 2010

### gabbagabbahey

Careful, is the charged enclosed by your Gaussian cylinder really $\lambda l$ always? ...Even if $r<3.6 \text{cm}$?

7. Jan 31, 2010

### LonghornDude8

Not if you're inside, but the radius I'm given is 1.9 cm and the diameter is 3.6 cm (1.8 cm radius). Therefore, the point it's asking for is outside the inner cylinder

8. Jan 31, 2010

### gabbagabbahey

Hmmm..yes, I had thought 3.6 cm was the radius....in that case, I don't see the problem...I too get 4824.9 N/C....is there more to this problem that you aren't telling us?

9. Jan 31, 2010

### LonghornDude8

Short of scanning the image for the picture, that's it. I'm thinking it might be negative but I wanted to see if anyone could catch an error in what I did

10. Jan 31, 2010

### LonghornDude8

11. Jan 31, 2010

### gabbagabbahey

So, to be clear, are you telling me that the answer to the first part of the problem isn't 4824.9 N/C?
Or, are you having trouble with the other parts of the problem? (The ones you didn't mention in your original post )

12. Jan 31, 2010

### LonghornDude8

The answer is NOT 4824.9 N/C. I haven't tried the rest, I'm waiting until I get this one correct

Here's where I looked to see if my method of solving it was correct:

Last edited: Jan 31, 2010
13. Jan 31, 2010

### gabbagabbahey

What is the answer according to whatever solution manual you are using?...I'm 99.9% sure that 4824.9 N/C is the correct answer, so I'm guessing your solution manual is incorrect.

14. Jan 31, 2010

### LonghornDude8

It's an online homework thing from the University of Texas. The only way to know the correct answer is if you get the correct answer.

15. Jan 31, 2010

### gabbagabbahey

Sounds like it might just be a matter of rounding and sig digs....I hate those online submission things, they're soooo picky!

16. Jan 31, 2010

### LonghornDude8

They round to 6 sig digs & they accept any answer within a 1% error. You can change the answer by one and it'll still be within that 1% error range. Also they don't round until the end which is what I always do so I should get their exact answer. My guess is that the answer is either negative or the picture is misleading and the values given are the radius's not diameters

17. Jan 31, 2010

### LonghornDude8

ya, I tried zero and got it correct... That was such a bad picture

18. Jan 31, 2010

### LonghornDude8

Now I've ran into a new problem in the same set, here's the question:
What is the surface charge density inside the hollow cylinder? Answer in units of nC/m2
Because the field on the inside of the hollow cylinder is zero, the net charge must be zero (since the area is definitely not zero).
Setup: $$Q/l=-\lambda$$ and $$Q/A=\sigma$$
set them equal and do some rearranging and you get:
$$\sigma = -\lambda l/A$$
plug $$2\pi rl$$ in for A:
$$\sigma = -\lambda /2\pi r$$
solve it (5.1 nC/m is lambda, 8.2 cm is r) and you get:
-9.89866 nC/m2
however, that is incorrect