# Homework Help: Gauss' Law and electric fields in non-conducting cylinder

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?

21. Jul 18, 2008

### dragonrider

Yes but the only formula that I know on how to find qenclosed is qenc= [(2*π*l)*(α*r3)]/(3*ε0) and when I finally derive the equation I end up with the same as the one from part b.

22. Jul 18, 2008

### Dick

No, it doesn't have any net charge. It will have equal and opposite surface charges on the inside and outside, but they cancel. You should end up with the same formula as part b.

23. Jul 18, 2008

### dragonrider

So use the equation α*r3/(3*R*ε0) where r = 15.2 cm and R = 18.4cm?

24. Jul 18, 2008

### Dick

Why would you say r=15.2cm??! There is only charge out to a radius of 5.48cm.

25. Jul 18, 2008

### dragonrider

Oh! I got it. Thank you very much.