# Homework Help: Insulating Cylindrical Shell

1. Sep 15, 2007

### Winzer

1. The problem statement, all variables and given/known data A long cylindrical insulating shell has an inner radius of a = 1.38 m and an outer radius of b = 1.61 m. The shell has a constant charge density of 4.80 C/m3.
1. What is the magnitude of the electric field at a distance of 1.87 m from the axis?
2. What is the magnitude of the electric field at a distance of 1.51 m from the axis?
3. If we take the potential at the axis to be zero, what is the electric potential at the outer radius of the shell?

2. Relevant equations
$$E=\frac{kq}{r^2}$$
$$\oint \vec{E}\vec{dr}= \frac{Q_{Enclosed}}{\epsilon_{o}}$$

3. The attempt at a solution
For 1 i use $$E=\frac{kq}{r^2}$$ but i am confused on finding q because i am given density of c/m^3. I am assuming an infinity long cyclinder?$$\vec{}$$

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2. Sep 15, 2007

### learningphysics

careful.. the integral should be E.da, not E.dr

Take a cylindrical section of arbitrary length L...

3. Sep 15, 2007

### Winzer

You are correct dA.
So:
$$dE= \frac{kdQ}{r^2}, dQ=pdV \longrightarrow dQ= 4.80 C/m^3 * 2\pi r l dr$$

4. Sep 15, 2007

### Winzer

Then:
$$E= \int \frac{k 4.80* 2 \pi r l}{r^2}dr$$
With he boundries a=a b=b

5. Sep 15, 2007

### learningphysics

No... use gauss' law... what is your gaussian shell... what is the charge enclosed...

6. Sep 15, 2007

### Winzer

Oh of course..I would pick a cyclinder as my G surface.

7. Sep 15, 2007

### learningphysics

exactly. can you write out the integral using gauss law for the first part? we need the field 1.87 m away.

Last edited: Sep 15, 2007
8. Sep 15, 2007

### Winzer

Ok, thanks. i got 1 & 2.

9. Sep 15, 2007

### Winzer

I know $$V=\frac{kq}{r}$$

10. Sep 15, 2007

### learningphysics

Nah... I don't think that'll work...

The voltage at r minus the voltage at 0 is:

$$-\int_0^r{\vec{E}\cdot\vec{dr}$$

You need to integrate this from 0 to b=1.61...

11. Sep 15, 2007

### Winzer

But then i get a ln(r/0), with other stuff infront. ln is -inf?

12. Sep 15, 2007

### learningphysics

That doesn't look right... the integral from 0 to 1.38m is 0... because the field is 0 for this part... so you just need the integral from 1.38 to 1.61.

13. Sep 16, 2007

### Winzer

so does:
$$V=\frac{-p (b^2-a^2)}{2 \epsilon_{o}}ln(\frac{b}{a})$$
sound on the right track?

14. Sep 16, 2007

### learningphysics

Hmm... I'm getting something different. Can you show how you got that integral?

15. Sep 16, 2007

### Winzer

Well how should I do it?

16. Sep 16, 2007

### learningphysics

Get the field inside the insulator. Then do $$-\int_{1.38}^{1.61}\vec{E}\cdot\vec{dr}$$

What do you get for the field in terms of r?

Use gauss' law. $$4.80*{\pi}r^2h = 2{\pi}rh*E$$

Last edited: Sep 16, 2007
17. Sep 16, 2007

### Winzer

mmm.. i Think it should be $$E= \frac{-p(b^2-a^2)}{2r \epsilon_{o}}$$
I choose $$-p(b^2-a^2)=Q_{enclosed}$$ because i had to integrate to find the charge enclosed.

18. Sep 16, 2007

### learningphysics

But the charge enclosed changes with r. You're integrating from r = 1.38 to 1.61... as you increase r, the charge enclosed becomes bigger.

19. Sep 16, 2007

### Winzer

so b is replaced with r

20. Sep 16, 2007

### learningphysics

Ah... yes, you're right... I made a mistake in my post. I should have written:

$$4.80*{\pi}(r^2 - a^2)h = 2{\pi}rh*E$$

So that gives what you have by replacing b with r.