Insulating Cylindrical Shell

Winzer

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

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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learningphysics

Homework Helper
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
$$E=\frac{kq}{r^2}$$
$$\oint \vec{E}\vec{dr}= \frac{Q_{Enclosed}}{\epsilon_{o}}$$
careful.. the integral should be E.da, not E.dr

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?$$\vec{}$$
Take a cylindrical section of arbitrary length L...

Winzer

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

Take a cylindrical section of arbitrary length L...
You are correct dA.
So:
$$dE= \frac{kdQ}{r^2}, dQ=pdV \longrightarrow dQ= 4.80 C/m^3 * 2\pi r l dr$$

Winzer

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

learningphysics

Homework Helper
Then:
$$E= \int \frac{k 4.80* 2 \pi r l}{r^2}dr$$
With he boundries a=a b=b
No... use gauss' law... what is your gaussian shell... what is the charge enclosed...

Winzer

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

learningphysics

Homework Helper
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:

Winzer

Ok, thanks. i got 1 & 2.

Winzer

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

learningphysics

Homework Helper
I know $$V=\frac{kq}{r}$$
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...

Winzer

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

learningphysics

Homework Helper
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.

Winzer

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

learningphysics

Homework Helper
so does:
$$V=\frac{-p (b^2-a^2)}{2 \epsilon_{o}}ln(\frac{b}{a})$$
sound on the right track?
Hmm... I'm getting something different. Can you show how you got that integral?

Winzer

Well how should I do it?

learningphysics

Homework Helper
Well how should I do it?
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:

Winzer

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. $$4.80*{\pi}r^2h = 2{\pi}rh*E$$
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.

learningphysics

Homework Helper
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.
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.

Winzer

so b is replaced with r

learningphysics

Homework Helper
so b is replaced with r
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.

Winzer

ok, cool I got it. I just made the mistake of not letting charge vary with r.

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