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Gauss's Law with non-uniform E-field

by flyingpig
Tags: efield, gauss, nonuniform
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flyingpig
#1
Jul23-11, 05:09 AM
P: 2,568
1. The problem statement, all variables and given/known data

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

3. 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...
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Doc Al
#2
Jul23-11, 05:24 AM
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Quote Quote by flyingpig View Post
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.
flyingpig
#3
Jul23-11, 05:29 AM
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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?

Doc Al
#4
Jul23-11, 05:34 AM
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Gauss's Law with non-uniform E-field

Almost. It's a cylinder, not a sphere.
flyingpig
#5
Jul23-11, 06:17 AM
P: 2,568
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]
Doc Al
#6
Jul23-11, 06:28 AM
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Quote Quote by flyingpig View Post
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.
flyingpig
#7
Jul23-11, 06:49 AM
P: 2,568
Definite or indefinite integral? What are my limits?
Doc Al
#8
Jul23-11, 06:51 AM
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Quote Quote by flyingpig View Post
Definite or indefinite integral? What are my limits?
You want the total charge from 0 to r.
flyingpig
#9
Jul23-11, 06:54 AM
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Quote Quote by Doc Al View Post
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?
flyingpig
#10
Jul23-11, 09:47 AM
P: 2,568
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?
flyingpig
#11
Jul24-11, 09:58 AM
P: 2,568
So do I still need to take the ratios between the volumes?
I like Serena
#12
Jul24-11, 10:25 AM
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Hi flyingpig!

Quote Quote by flyingpig View Post
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.


Quote Quote by flyingpig View Post
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?



Quote Quote by flyingpig View Post
So do I still need to take the ratios between the volumes?
Which ratio?
Doc Al
#13
Jul24-11, 01:33 PM
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Quote Quote by flyingpig View Post
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.
flyingpig
#14
Jul25-11, 09:54 AM
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Quote Quote by I like Serena View Post
Hi flyingpig!



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?
[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
SammyS
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Jul25-11, 12:45 PM
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Quote Quote by flyingpig View Post
...
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.
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Jul25-11, 01:42 PM
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Quote Quote by flyingpig View Post
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.


Quote Quote by flyingpig View Post
[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.
flyingpig
#17
Jul27-11, 06:26 AM
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Quote Quote by I like Serena View Post
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]
Doc Al
#18
Jul27-11, 07:25 AM
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Quote Quote by flyingpig View Post
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]


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

No ratios.


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