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

• flyingpig
In summary: 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}Good. (If you want to express it as a vector, include a unit vector to show the direction.)In summary, the electric field from the center of the axis is zero for a long cylindrical object with a non-uniform charge density.
flyingpig

## Homework Statement

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

## 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...

flyingpig said:
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.

$$V = \frac{4 \pi r^3}{3}$$

$$dV = 4\pi r^2 dr$$

$$\rho dV = e^{ar} 4\pi r^2 dr$$

Integrate that? This is just an indefinite integral right?

Almost. It's a cylinder, not a sphere.

Oh wait, what am i doing lol

$$V = \pi r^2 l$$

$$dV = 2\pi rl dr$$

$$\rho dV = e^{ar} 2\pi rl dr$$

flyingpig said:
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.

Definite or indefinite integral? What are my limits?

flyingpig said:
Definite or indefinite integral? What are my limits?
You want the total charge from 0 to r.

Doc Al said:
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?

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?

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

Hi flyingpig!

flyingpig said:
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.
flyingpig said:
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?
flyingpig said:
So do I still need to take the ratios between the volumes?

Which ratio?

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flyingpig said:
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.

I like Serena said:
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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flyingpig said:
...
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 canceled 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.

flyingpig said:
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.
flyingpig said:
$$\frac{Q_{en}}{Q_{charge\;of\;cylinder}}$$

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.

I like Serena said:
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

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

flyingpig said:
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.

Say the density was uniform, then we need ratios. So why don't we have ratios here??

flyingpig said:
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?

Because the charge enclosed inside the cylinder has a smaller Gaussan volume

flyingpig said:
Because the charge enclosed inside the cylinder has a smaller Gaussan volume
So?...

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.

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

Last edited by a moderator:
The volume ratio works for this example because the charge density is uniform throughout the sphere.

Why does being non-uniform make it any different?

Is there no meaning in this?

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

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flyingpig said:
Why does being non-uniform make it any different?
As I said earlier, the charge is always:

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

When the charge density is constant you can pull it out of the integral and then use the ratio of volumes and the total charge, if you like. (But why?)

When the charge density is not constant, you cannot do that.

Is there no meaning in this?

$$\frac{Q_{charge\;of\;cylinder}\int \rho dV}{\rho V} = \frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{ar} \pi R^2 l}= Q_{en}$$
It seems quite convoluted. (Note that, since the charge density is not constant, $\rho V$ doesn't mean what I assume you think it means.)

Sorry at the end i meant

$$\frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{aR} \pi R^2}$$

Like I evalutated the non-uniform density at R for e^ar

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Okay fixed! thanks

flyingpig said:
Sorry at the end i meant

$$\frac{Q_{charge\;of\;cylinder}2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{aR} \pi R^2}$$

Like I evalutated the non-uniform density at R for e^ar
What's that a calculation of? And why would your answer have Q in it--you are given the charge density as a function of a and r. I would not accept an answer that had Q in it.

This stuff $$\frac{[2\pi l \frac{e^{ar} (ar - 1) + 1}{a^2}}{e^{aR} \pi R^2$$

is supposed to cancel itself out (dimension analytically)

Just wondering, does that mean part ii will be the same answer?

flyingpig said:
Just wondering, does that mean part ii will be the same answer?
No. Why would you think that?

Because the gaussian surface is now bigger...?

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