Electric Field of an Insulating Cylindrical Shell

[Winzer]

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?

Homework Equations

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

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{}

 

[learningphysics]

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?

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

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]

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]

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.

 

[Winzer]

Ok, thanks. i got 1 & 2.

 

[Winzer]

how about 3?
I know V=\frac{kq}{r}

 

[learningphysics]

how about 3?
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]

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]

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]

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

 

[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]

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]

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.