# Gauss's Law with non-uniform E-field

by flyingpig
Tags: efield, gauss, nonuniform
 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 $$\rho = e^{ar}$$ 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) $$\oint \vec{E} \cdot d\vec{A} = \frac{\sum Q_{en}}{\epsilon_0}$$ $$\vec{E} 2\pi rl = \frac{\sum Q_{en}}{\epsilon_0}$$ So now here is the problem, if it is inside the cylinder I get something like (1) $$\rho V = Q$$ (2) $$\rho V' = Q_{en}$$ Divide them out and some algebra and I get $$Q\frac{V'}{V} = Q_{en}$$ 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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P: 41,459
 Quote by flyingpig So now here is the problem, if it is inside the cylinder I get something like (1) $$\rho V = Q$$
You must integrate to find the total charge within your Gaussian surface.
 P: 2,568 $$V = \frac{4 \pi r^3}{3}$$ $$dV = 4\pi r^2 dr$$ $$\rho dV = e^{ar} 4\pi r^2 dr$$ Integrate that? This is just an indefinite integral right?
 Mentor P: 41,459 Gauss's Law with non-uniform E-field Almost. It's a cylinder, not a sphere.
 P: 2,568 Oh wait, what am i doing lol $$V = \pi r^2 l$$ $$dV = 2\pi rl dr$$ $$\rho dV = e^{ar} 2\pi rl dr$$
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P: 41,459
 Quote by flyingpig Oh wait, what am i doing lol $$V = \pi r^2 l$$ $$dV = 2\pi rl dr$$ $$\rho dV = e^{ar} 2\pi rl dr$$
Good.
 P: 2,568 Definite or indefinite integral? What are my limits?
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P: 41,459
 Quote by flyingpig Definite or indefinite integral? What are my limits?
You want the total charge from 0 to r.
P: 2,568
 Quote by Doc Al 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 $$\rho = \frac{1}{r}$$

What would the discontinuity mean? Would it mean that there is no Electric field in the line of its axis?
 P: 2,568 So anyways... $$2\pi l\int_{0}^{r} re^{ar} dr = 2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}$$ $$\vec{E} = \frac{1}{\epsilon_0 r}\frac{e^{ar} (ar - 1) + 1}{a^2}$$ So how do I tell the direction...? It looks all positive, it radiates outward?
 P: 2,568 So do I still need to take the ratios between the volumes?
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P: 6,187
Hi flyingpig!

 Quote by flyingpig What happens if $$\rho = \frac{1}{r}$$ 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 by flyingpig So anyways... $$2\pi l\int_{0}^{r} re^{ar} dr = 2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}$$ $$\vec{E} = \frac{1}{\epsilon_0 r}\frac{e^{ar} (ar - 1) + 1}{a^2}$$ 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 by flyingpig So do I still need to take the ratios between the volumes?
Which ratio?
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P: 41,459
 Quote by flyingpig So anyways... $$2\pi l\int_{0}^{r} re^{ar} dr = 2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}$$
Looks good. That's the charge.

 $$\vec{E} = \frac{1}{\epsilon_0 r}\frac{e^{ar} (ar - 1) + 1}{a^2}$$
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.
P: 2,568
 Quote by I like Serena 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 $$\rho(0) = c$$ and treat it as a piecewise function?

 Which ratio?
$$\frac{Q_{en}}{Q_{charge\;of\;cylinder}}$$

$$\frac{\int \rho dV}{\rho V} = \frac{Q_{en}}{Q_{charge of cylinder}}$$

$$\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = Q_{en}$$

For $$\rhoV = e^{ar} 2\pi Rl$$

Where R is the radius of the cylinder
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PF Gold
P: 7,808
 Quote by flyingpig ... What happens if $$\rho = \frac{1}{r}$$ 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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P: 6,187
 Quote by flyingpig What happens if I say $$\rho(0) = c$$ 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 by flyingpig $$\frac{Q_{en}}{Q_{charge\;of\;cylinder}}$$ 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.
P: 2,568
 Quote by I like Serena 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

$$\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = \frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{ar} \pi R^2 l}= Q_{en}$$
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P: 41,459
 Quote by flyingpig Yeah it does, for an insulating surface with uniform density. I was thinking of $$\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = \frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{ar} \pi R^2 l}= Q_{en}$$

$$Q_{en} = \int \rho dV$$

No ratios.

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