
#1
Jul2311, 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 nonuniform 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 nonuniform density? I will stop here before I do ii... 



#2
Jul2311, 05:24 AM

Mentor
P: 40,872





#3
Jul2311, 05:29 AM

P: 2,568

[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
Jul2311, 05:34 AM

Mentor
P: 40,872

Gauss's Law with nonuniform Efield
Almost. It's a cylinder, not a sphere.




#5
Jul2311, 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] 



#7
Jul2311, 06:49 AM

P: 2,568

Definite or indefinite integral? What are my limits?




#9
Jul2311, 06:54 AM

P: 2,568

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
Jul2311, 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? 



#11
Jul2411, 09:58 AM

P: 2,568

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




#12
Jul2411, 10:25 AM

HW Helper
P: 6,189

Hi flyingpig!
That's not physically possible. 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? 



#13
Jul2411, 01:33 PM

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P: 40,872





#14
Jul2511, 09:54 AM

P: 2,568

[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 



#15
Jul2511, 12:45 PM

Emeritus
Sci Advisor
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PF Gold
P: 7,379





#16
Jul2511, 01:42 PM

HW Helper
P: 6,189

Still physically impossible. Sorry. 



#17
Jul2711, 06:26 AM

P: 2,568

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] 


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