Gauss' Law and electric fields in non-conducting cylinder

[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.

 

[Dick]

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

 

[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

 

[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.

 

[dragonrider]

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

 

[Dick]

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

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

 

[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.

 

[Dick]

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.

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??

 

[dragonrider]

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

 

[Dick]

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

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.

 

[dragonrider]

Here is a link to it.

http://img155.imageshack.us/img155/662/problem1qq1.th.gif

 

[Dick]

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

 

[dragonrider]

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

How about the link can you see using that?

 

[Dick]

How about the link can you see using that?

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.

 

[dragonrider]

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

 

[Dick]

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

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

 

[dragonrider]

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

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.

 

[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.

 

[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?

 

[Dick]

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?

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

 

[dragonrider]

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

Yes but the only formula that I know on how to find q_enclosed is q_enc= [(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.

 

[Dick]

Yes but the only formula that I know on how to find q_enclosed is q_enc= [(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.

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.

 

[dragonrider]

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

 

[Dick]

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

 

[dragonrider]

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

Oh! I got it. Thank you very much.