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Insulating Cylindrical Shell

  • Thread starter Winzer
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1. Homework Statement 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. Homework Equations
[tex] E=\frac{kq}{r^2}[/tex]
[tex] \oint \vec{E}\vec{dr}= \frac{Q_{Enclosed}}{\epsilon_{o}}[/tex]

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

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learningphysics

Homework Helper
4,099
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1. Homework Statement 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. Homework Equations
[tex] E=\frac{kq}{r^2}[/tex]
[tex] \oint \vec{E}\vec{dr}= \frac{Q_{Enclosed}}{\epsilon_{o}}[/tex]
careful.. the integral should be E.da, not E.dr

3. The Attempt at a Solution
For 1 i use [tex] E=\frac{kq}{r^2}[/tex] but i am confused on finding q because i am given density of c/m^3. I am assuming an infinity long cyclinder?[tex]\vec{}[/tex]
Take a cylindrical section of arbitrary length L...
 
598
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careful.. the integral should be E.da, not E.dr

Take a cylindrical section of arbitrary length L...
You are correct dA.
So:
[tex] dE= \frac{kdQ}{r^2}, dQ=pdV \longrightarrow dQ= 4.80 C/m^3 * 2\pi r l dr[/tex]
 
598
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Then:
[tex] E= \int \frac{k 4.80* 2 \pi r l}{r^2}dr[/tex]
With he boundries a=a b=b
 

learningphysics

Homework Helper
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Then:
[tex] E= \int \frac{k 4.80* 2 \pi r l}{r^2}dr[/tex]
With he boundries a=a b=b
No... use gauss' law... what is your gaussian shell... what is the charge enclosed...
 
598
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Oh of course..I would pick a cyclinder as my G surface.
 

learningphysics

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Oh of course..I would pick a cyclinder as my G surface.
exactly. can you write out the integral using gauss law for the first part? we need the field 1.87 m away.
 
Last edited:
598
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Ok, thanks. i got 1 & 2.
 
598
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how about 3?
I know [tex] V=\frac{kq}{r}[/tex]
 

learningphysics

Homework Helper
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how about 3?
I know [tex] V=\frac{kq}{r}[/tex]
Nah... I don't think that'll work...

The voltage at r minus the voltage at 0 is:

[tex]-\int_0^r{\vec{E}\cdot\vec{dr}[/tex]

You need to integrate this from 0 to b=1.61...
 
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But then i get a ln(r/0), with other stuff infront. ln is -inf?
 

learningphysics

Homework Helper
4,099
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But then i get a ln(r/0), with other stuff infront. ln is -inf?
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.
 
598
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so does:
[tex] V=\frac{-p (b^2-a^2)}{2 \epsilon_{o}}ln(\frac{b}{a})[/tex]
sound on the right track?
 

learningphysics

Homework Helper
4,099
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so does:
[tex] V=\frac{-p (b^2-a^2)}{2 \epsilon_{o}}ln(\frac{b}{a})[/tex]
sound on the right track?
Hmm... I'm getting something different. Can you show how you got that integral?
 
598
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Well how should I do it?
 

learningphysics

Homework Helper
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Well how should I do it?
Get the field inside the insulator. Then do [tex]-\int_{1.38}^{1.61}\vec{E}\cdot\vec{dr}[/tex]

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

Use gauss' law. [tex]4.80*{\pi}r^2h = 2{\pi}rh*E[/tex]
 
Last edited:
598
0
Get the field inside the insulator. Then do [tex]-\int_{1.38}^{1.61}\vec{E}\cdot\vec{dr}[/tex]

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

Use gauss' law. [tex]4.80*{\pi}r^2h = 2{\pi}rh*E[/tex]
mmm.. i Think it should be [tex]E= \frac{-p(b^2-a^2)}{2r \epsilon_{o}}[/tex]
I choose [tex] -p(b^2-a^2)=Q_{enclosed}[/tex] because i had to integrate to find the charge enclosed.
 

learningphysics

Homework Helper
4,099
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mmm.. i Think it should be [tex]E= \frac{-p(b^2-a^2)}{2r \epsilon_{o}}[/tex]
I choose [tex] -p(b^2-a^2)=Q_{enclosed}[/tex] because i had to integrate to find the charge enclosed.
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.
 
598
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so b is replaced with r
 

learningphysics

Homework Helper
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so b is replaced with r
Ah... yes, you're right... I made a mistake in my post. I should have written:

[tex]4.80*{\pi}(r^2 - a^2)h = 2{\pi}rh*E[/tex]

So that gives what you have by replacing b with r.
 
598
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ok, cool I got it. I just made the mistake of not letting charge vary with r.
 

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