Gauss's Law with non-uniform E-field

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Homework Statement



A long insulating cylinder has radius R, length l, and a non-uniform charge density per volume [tex]\rho = e^{ar}[/tex] where r is the distance from the axis of the cylinder. Find the electric field from the center of the axis for

i) r < R
ii) r > R

The Attempt at a Solution



i)

[tex]\oint \vec{E} \cdot d\vec{A} = \frac{\sum Q_{en}}{\epsilon_0}[/tex]


[tex] \vec{E} 2\pi rl = \frac{\sum Q_{en}}{\epsilon_0}[/tex]

So now here is the problem, if it is inside the cylinder I get something like

(1) [tex]\rho V = Q[/tex]
(2) [tex]\rho V' = Q_{en}[/tex]

Divide them out and some algebra and I get

[tex]Q\frac{V'}{V} = Q_{en}[/tex]

Should I keep this? Does it even matter if it was a non-uniform density?

I will stop here before I do ii...
 

Answers and Replies

  • #2
Doc Al
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So now here is the problem, if it is inside the cylinder I get something like

(1) [tex]\rho V = Q[/tex]
You must integrate to find the total charge within your Gaussian surface.
 
  • #3
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[tex]V = \frac{4 \pi r^3}{3}[/tex]

[tex]dV = 4\pi r^2 dr[/tex]

[tex]\rho dV = e^{ar} 4\pi r^2 dr[/tex]

Integrate that? This is just an indefinite integral right?
 
  • #4
Doc Al
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Almost. It's a cylinder, not a sphere.
 
  • #5
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Oh wait, what am i doing lol

[tex]V = \pi r^2 l [/tex]

[tex]dV = 2\pi rl dr[/tex]

[tex]\rho dV = e^{ar} 2\pi rl dr[/tex]
 
  • #6
Doc Al
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Oh wait, what am i doing lol

[tex]V = \pi r^2 l [/tex]

[tex]dV = 2\pi rl dr[/tex]

[tex]\rho dV = e^{ar} 2\pi rl dr[/tex]
Good.
 
  • #7
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Definite or indefinite integral? What are my limits?
 
  • #8
Doc Al
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Definite or indefinite integral? What are my limits?
You want the total charge from 0 to r.
 
  • #9
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You want the total charge from 0 to r.

I had a feeling it was going to be a "0" to something

What happens if [tex]\rho = \frac{1}{r}[/tex]

What would the discontinuity mean? Would it mean that there is no Electric field in the line of its axis?
 
  • #10
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So anyways...

[tex]2\pi l\int_{0}^{r} re^{ar} dr = 2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}[/tex]

[tex]\vec{E} = \frac{1}{\epsilon_0 r}\frac{e^{ar} (ar - 1) + 1}{a^2}[/tex]

So how do I tell the direction...? It looks all positive, it radiates outward?
 
  • #11
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So do I still need to take the ratios between the volumes?
 
  • #12
I like Serena
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Hi flyingpig! :smile:

What happens if [tex]\rho = \frac{1}{r}[/tex]

What would the discontinuity mean? Would it mean that there is no Electric field in the line of its axis?

Such a distribution would mean you have infinite charge at the axis of the cylinder.
That's not physically possible.


So anyways...

[tex]2\pi l\int_{0}^{r} re^{ar} dr = 2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}[/tex]

[tex]\vec{E} = \frac{1}{\epsilon_0 r}\frac{e^{ar} (ar - 1) + 1}{a^2}[/tex]

So how do I tell the direction...? It looks all positive, it radiates outward?

You did not calculate the vector E, so you shouldn't write it down that way.
You only calculated the component of E that is perpendicular to the surface that you integrated, which is the radial component.
Furthermore you have assumed that it is the same everywhere on the surface of the cylinder.
This is a reasonable assumption for a long cylinder with a symmetric distribution of charge.

There are 2 more components to the E field at any point.
Typically you would express them in cylindrical coordinates, meaning you have a component along the length of the cylinder, and a tangential component in the direction of the angle.

Can you deduce what these components are?



So do I still need to take the ratios between the volumes?

Which ratio? :confused:
 
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  • #13
Doc Al
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So anyways...

[tex]2\pi l\int_{0}^{r} re^{ar} dr = 2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}[/tex]
Looks good. That's the charge.

[tex]\vec{E} = \frac{1}{\epsilon_0 r}\frac{e^{ar} (ar - 1) + 1}{a^2}[/tex]
Good. (If you want to express it as a vector, include a unit vector to show the direction.)

So how do I tell the direction...? It looks all positive, it radiates outward?
Yes. Assuming the charge is positive, the field points outward.
 
  • #14
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Hi flyingpig! :smile:



Such a distribution would mean you have infinite charge at the axis of the cylinder.
That's not physically possible.

What happens if I say [tex]\rho(0) = c[/tex] and treat it as a piecewise function?


Which ratio? :confused:

[tex]\frac{Q_{en}}{Q_{charge\;of\;cylinder}}[/tex]

[tex]\frac{\int \rho dV}{\rho V} = \frac{Q_{en}}{Q_{charge of cylinder}}[/tex]

[tex]\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = Q_{en}[/tex]

For [tex]\rhoV = e^{ar} 2\pi Rl[/tex]

Where R is the radius of the cylinder
 
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  • #15
SammyS
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...
What happens if [tex]\rho = \frac{1}{r}[/tex]

What would the discontinuity mean?
dV is proportional to r, so the effect of 1/r would be cancelled out when you integrate to find the charge..
Would it mean that there is no Electric field in the line of its axis?
Symmetry shows that there E=0 along the axis.
 
  • #16
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What happens if I say [tex]\rho(0) = c[/tex] and treat it as a piecewise function?

Well, with a function like 1/r the charge will still be near infinity if you get close enough to zero.
Still physically impossible.
Sorry.


[tex]\frac{Q_{en}}{Q_{charge\;of\;cylinder}}[/tex]

Where R is the radius of the cylinder

Errr... no, Gauss's formula for electric field and enclosed charge does not involve such a ratio.
 
  • #17
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Errr... no, Gauss's formula for electric field and enclosed charge does not involve such a ratio.

Yeah it does, for an insulating surface with uniform density.

I was thinking of

[tex]\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = \frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{ar} \pi R^2 l}= Q_{en}[/tex]
 
  • #18
Doc Al
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Yeah it does, for an insulating surface with uniform density.

I was thinking of

[tex]\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = \frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{ar} \pi R^2 l}= Q_{en}[/tex]
:confused:

[tex]Q_{en} = \int \rho dV[/tex]

No ratios.
 
  • #19
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Say the density was uniform, then we need ratios. So why don't we have ratios here??
 
  • #20
Doc Al
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Say the density was uniform, then we need ratios. So why don't we have ratios here??
If you know the density, why do you need a ratio?
 
  • #21
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Because the charge enclosed inside the cylinder has a smaller Gaussan volume
 
  • #22
Doc Al
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Because the charge enclosed inside the cylinder has a smaller Gaussan volume
So?...
 
  • #23
Doc Al
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What you're probably thinking is that if you know the total charge per length of the cylinder, then you can use a ratio of volumes to find the charge enclosed in a Gaussian surface within the cylinder.

Qenclosed = Qtotal (Volume enclosed/Volume total)

But not necessary if you know the density.
 
  • #24
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OKay i'll just pull an example from where I got this idea of volume ratios

[PLAIN]http://img585.imageshack.us/img585/1779/unled2jp.jpg [Broken]
 
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  • #25
SammyS
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The volume ratio works for this example because the charge density is uniform throughout the sphere.
 

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