# Gauss Law/Electric Potential Question(s)

1. Oct 1, 2012

### Kalookakoo

1. The problem statement, all variables and given/known data

http://ft.trillian.im/b0f9eae6eee2cb82091d0bd460ed09948b0cf819/6b66r5cGJrRSOCQPbItbRmtMARJhR.jpg

2. Relevant equations
http://ft.trillian.im/b0f9eae6eee2cb82091d0bd460ed09948b0cf819/6b674JdPpi29mGe8SqWkcnh8K6bmA.jpg

3. The attempt at a solution

I know how it works here. I will not be spoonfed or given answers, which is why I'm a little skeptical I'll actually receive help with this problem lol.

I've been having problems with these kinda of questions for awhile now, and I have a test next week. If someone could just hint me the thought process behind questions like this so I can slowly go through this (this assignment is due thursday morning) that would be great. My professor knows what he lectures about, but hes hard to comprehend and he doesn't explain much. There isn't enough personal attention for the tutoring (thats what it is now I guess) I need in this course...

I think #2 is the easier of the two, so why not start there? The electric potential when r<a, or otherwise said as within that inner cylinder. So I want to find the electric field within that cylinder first I suppose?

It's so frustrating how far I've fallen behind compared to my other classes I'm kicking butt in...

2. Oct 1, 2012

### BruceW

these types of questions need lots of practice, and the answers are not always totally intuitive, so it is sometimes difficult to see what the answer should be. But anyway, question #2 looks like a good one to start practice with. You've mentioned Gauss' law in the title, and that is what you need. So start out by using a form of Gauss' law which will hopefully take advantage of the symmetry in the problem.

3. Oct 1, 2012

### Kalookakoo

Ok So...

∫E*da = q/E0
EA = q/AE0
E = q/(2∏rLE)0

Q enclosed isn't given so I assume I get that from the charge per unit length. (charge per unit length = q/L)
I will use T as charge per unit length.
Q = T*L

So:
E = T/(2∏rE0)

Hopefully I'm not doing unnecessary work...from here...
ΔV = ∫E*dl
V = E*L

So V = (T*L)/(2∏rE0)

Is that right for r<a?

4. Oct 2, 2012

### BruceW

Exactly! You are actually pretty good at this stuff.

Um, be careful about the direction you are taking the line integral. Think of which direction the electric field will be, and so which direction must you take the line integral to get V.

5. Oct 2, 2012

### Kalookakoo

Well, I feel like I have all the information and formula in my head somewhere. Just need some guidance sorting it all out. Haha

So, upon further review, I see that the line integral should be negative. I can't conceptually understand why though. The field is going from the inner cylinder (+) to the outer cylinder (-), so isn't the the proper direction?

I'll take a crack at b (and maybe more?) in between lectures. Have to get to class now!

6. Oct 2, 2012

### BruceW

You are correct that the electric field is going from the inner cylinder to the outer cylinder, but you are not integrating correctly. You have assumed that E does not vary with respect to this direction, but this is not true. Also, the equation ΔV = ∫E*dl is not quite right.

7. Oct 2, 2012

### Kalookakoo

Is ΔV = ∫E*dr correct? Wikipedia has it with a negative, but I can't seem to understand why.

I realize now everything with E is constant except for r.

V = T/(2∏E0) * ∫(1/r)dr
V = T/(2∏E0) * ln(r)*r

So V = [Tr/(2∏E0)]*ln(r)

b) a < r < b

This is between the two cylinders. I'm going to assume we have to take the seperate electric fields and add them together and use the answer? Finding the respective electric potentials

Inner Cylinder:
I just use what I had before?
E = T/(2∏rE0)

Outer Cylinder:
Hmm. I have a problem. Do I use r for this as well? or another distance from this radius.

8. Oct 3, 2012

### BruceW

It must have the negative sign. As an example, Imagine there is some big charge, which creates a positive potential around it. Which direction will the electric field be? And how does the potential change as you go further away from the big charge? This should show you why there must be the negative sign.

You are right that E depends only on r. But the integration is not correct. Also, you have forgotten the Δ sign with the V. (Since the V will generally be non-zero at either end of the integral). And there is also the issue with the negative sign.

eh? You have already been trying to work out the potential in the bit from a<r<b. Remember, that you used charge enclosed=TL So you have assumed that r>a. If you wanted to find the potential for r<a, the charge enclosed would need to depend on r.

9. Oct 3, 2012

### Kalookakoo

Gotcha on the first part. thanks

I don't quite understand. So Q = TL can only be used for between the cylinders?

So when r<a, otherwise known as within the inner cylinder, it must depend on r? So I replace L with r? Maybe I have the wrong understanding of what exactly represents..ughhh.

I guess we shall carry on with a<r<b for now.

so
ΔV = T/(2∏E0) * -∫(1/r)dr
The integral is from a to b?
So....
V = T/(2∏E0) * -ln(b/a)?

Is it correct I should find the electrical potential from the inner cylinder and add it to the electric potential from the outer cylinder?

And how does it being an insulator come into play? (the outer cylinder)

10. Oct 3, 2012

### BruceW

You're right that Q=TL can only be used between the cylinders. And for inside the inner cylinder, you don't replace L with r. But the enclosed charge definitely does depend on r when r<a. Think about the 'Gaussian surface' which you are using, it is a cylindrical shape, right, and for r<a, your 'Gaussian surface' only encloses the charge up to r in the radial direction, and up to L in the z direction.

Yep, this is pretty close. Again it should be a change in V, since you haven't specified a V=0 point yet. If you look in the question, it actually tells you to where to have V=0. (Usually, we take V=0 at r=infinity, but this problem has explicitly asked you to do differently, which is fine).

Are you reading the bit in the question, where it hints that the net potential is the potential due to the sum of the conductors? Well, keep doing the problem the way you are. You are still calculating the potential for r<b, so for now you don't need to worry about the outer cylinder, since that charge is not enclosed by your 'Gaussian surface'.

I don't think it is an insulator. The question just says the stand which they both rest on is insulating. Can you think why they make sure that there is no conductor between the two cylinders?

11. Oct 3, 2012

### Kalookakoo

First off let me just say you've been a big help, and thank you for taking time out to help me understand this stuff.

Just gonna throw this out. Q = (TL)/R It makes sense in my head but it's just a guess.

oooooh. Didn't look down there.
V = 0 @ r=b
Change in V = V - where v = 0 ?
so...
V = [T/(2∏E0) * -ln(b/a)] - [T/(2∏E0) * -ln(b)]

or when I integrated in the first part, b being 0....so..
V = [T/(2∏E0) *-ln(a)]?

So time for the outer cylinder I guess.
I'm integrating from b to a....does my line integral become positive then?

V = [-T/(2∏E0) * ln(a/b)] - [T/(2∏E0) * -ln(b)]

I can't read or do simple math. I will be a great engineer...ha anyways....

Call me crazy but I believe the charge on a conductor rests on its surface, and the there is no field (or electric potential?) inside of the conductor...

I'm a conceptual guy...once math gets involved I do the oddest of things...

Last edited: Oct 3, 2012
12. Oct 4, 2012

### BruceW

Actually, I just realised that the charge inside the cylinder is very simple, something which you said made me realise this.
You have given yourself the answer here to the charge inside the cylinder. Although I should say, it's not necessarily true that there is no potential inside a conductor.

You are right that V=0 at r=b, but your equations from there don't make sense. Go back to your equation ΔV = T/(2∏E0) * -ln(b/a) This is correct (where ΔV = V(b)-V(a), if you've forgotten). Now, you've also got the information that V(b)=0, so you can use this.