# Gauss' Law and electric fields

1. Jul 18, 2008

### dragonrider

A long non-conducting cylinder has a charge density ρ = α*r, where α = 4.95 C/m4 and r is in meters. Concentric around it is a hollow metallic cylindrical shell.

https://www.physicsforums.com/attachment.php?attachmentid=14741&d=1216404841

1. What is the electric field at 2 cm from the central axis? Assume the length L is very long compared to the diameter of the shell, and neglect edge effects. Answer in units of N/C.
- I used the Gauss' Law and derived the equation α*r3/(3*ε0). And got the answer 7.454e+07 N/C and it was right.

2. What is the electric field at 5.84 cm from the central axis? Answer in units of N/C.
- But when I use the same equation here I get the wrong answer. Can some one help me please!

4. What is the electric field at 18.4 cm from the central axis? Answer in units of N/C.

2. Jul 18, 2008

### Dick

You mean α*r^2/(3*ε0), right? Do they say anything about the diameter of the shell?

3. Jul 18, 2008

### dragonrider

Yes, sorry about the mistake, the formula I found was α*r^2/(3*ε0). No the question doesn't say anything about the diameter of the expect the values given in the drawing.
https://www.physicsforums.com/attachment.php?attachmentid=14741&d=1216404841

4. Jul 18, 2008

### Dick

Is the value in the drawing? Your formula is only valid if r is inside of the cylinder. If it's outside, you'll need to get the charge by integrating over the whole cylinder and then applying Gauss' law.

5. Jul 18, 2008

### dragonrider

For part b of the question the value is between in the inner cylinder and the outer shell

6. Jul 18, 2008

### Dick

Then when you are integrating to get the charge, only integrate out to the radius of cylinder. Not all of the way to r.

7. Jul 18, 2008

### dragonrider

Do you mean just use r = 5.48 cm in the equation α*r^2/(3*ε0) because I did that and I still got it wrong.

8. Jul 18, 2008

### Dick

No, no, no. You will get a different formula if you apply Gauss' law when r is greater than the radius of the charge. What is the radius of the charged cylinder??

9. Jul 18, 2008

### dragonrider

radius of the charged cylinder from the figure is 15.2 cm for outer radius and 10.5 cm for inner radius

10. Jul 18, 2008

### Dick

I wish I could see your figure. I'm very confused. Are you sure those aren't diameters? You said for part b) that 5.84cm was outside of the inner cylinder.

11. Jul 18, 2008

### dragonrider

Last edited by a moderator: May 3, 2017
12. Jul 18, 2008

### Dick

Doh. Now it has to be approved. Hope you're not in a hurry.

13. Jul 18, 2008

### dragonrider

14. Jul 18, 2008

### Dick

Yes, thanks. Ok, so when you use Gauss' law, to compute the charge only integrate out to 5.48cm. When you compute E*Area, use r=5.84cm.

15. Jul 18, 2008

### dragonrider

So you want me compute E using the equation α*r3/(3*R*ε0) where r = 5.48 cm and R = 5.84cm?

16. Jul 18, 2008

### Dick

That seems right, doesn't it? It's Gauss' law. Can't go wrong with that.

17. Jul 18, 2008

### dragonrider

It still says it is wrong am I supposed to cube or square the small r. Because when I used the equation that I showed you and cubed the small r and got 5.251e+08 but it was still wrong.

Last edited: Jul 18, 2008
18. Jul 18, 2008

### Dick

I cubed the small r and used that formula and I got a different number. Be sure all of your lengths are in meters.

19. Jul 18, 2008

### dragonrider

Oh well I just ran out all my tries for that part but how about the last part when it's asking for E at 18.4 cm do I use the same equation or does it become different since it's outside the shell?

20. Jul 18, 2008

### Dick

You tell me. You still use Gauss' law. Does the shell carry any net charge?