Gauss's Law with non-uniform E-field

[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&#039; = Q_{en}

Divide them out and some algebra and I get

Q\frac{V&#039;}{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...

 

[Doc Al]

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.

 

[flyingpig]

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?

 

[Doc Al]

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

 

[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

 

[Doc Al]

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.

 

[flyingpig]

Definite or indefinite integral? What are my limits?

 

[Doc Al]

Definite or indefinite integral? What are my limits?

You want the total charge from 0 to r.

 

[flyingpig]

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?

 

[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?

 

[flyingpig]

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

 

[I like Serena]

Hi 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.

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?

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

Which ratio?

 

[Doc Al]

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.

 

[flyingpig]

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

 

[SammyS]

...
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.

 

[I like Serena]

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.

\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.

 

[flyingpig]

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}

 

[Doc Al]

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.

 

[flyingpig]

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

 

[Doc Al]

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?

 

[flyingpig]

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

 

[Doc Al]

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

So?...

 

[Doc Al]

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.

 

[flyingpig]

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

 

[SammyS]

The volume ratio works for this example because the charge density is uniform throughout the sphere.

 

[flyingpig]

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}

 

[Doc Al]

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.)

 

[flyingpig]

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

 

[Doc Al]

(Check your use of braces in your latex expression)

 

[flyingpig]

Okay fixed! thanks

 

[Doc Al]

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.

 

[flyingpig]

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)

 

[flyingpig]

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

 

[Doc Al]

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

No. Why would you think that?

 

[flyingpig]

Because the gaussian surface is now bigger...?

 

[flyingpig]

Actually a better question would be, why wouldn't it be different?

 

[Doc Al]

When inside the charged sphere, as r changes so does the total charge contained within r. But once you are outside of the sphere, the total charge remains fixed.

 

[flyingpig]

Why? Shouldn't it get weaker?

 

[flyingpig]

Someone come back

 

[Doc Al]

Why? Shouldn't it get weaker?

Shouldn't what get weaker? (If you're responding to my last post, realize that I was talking about charge not field strength.)

 

[flyingpig]

Shouldn't what get weaker? (If you're responding to my last post, realize that I was talking about charge not field strength.)

Well the density is non-uniform

\rho = \frac{Q}{V}

Volume can't really change so

\rho\; \alpha \;Q

so as radius goes up, density goes down and since they are proportional, charge drops?

 

[Doc Al]

Well the density is non-uniform

\rho = \frac{Q}{V}

Volume can't really change so

\rho\; \alpha \;Q

so as radius goes up, density goes down and since they are proportional, charge drops?

I have no idea what you're talking about.

Within the sphere, obviously the total charge within your Gaussian sphere increases with radius; Outside the sphere, there's no additional charge so the total charge remains fixed.

 

[flyingpig]

Q_{en} = \int \rho dV

I am going to take a while shot at this one. Since we are looking at r > R, that means we only care about the field r > R

I think it would be

Q_{en} = \int_{R}^{r} \rho dV

But r > R not greater or EQUAL to. In other words, even though our Gaussian surface encloses the whole surface, it only cares about outside of r > R

 

[Doc Al]

I think it would be

Q_{en} = \int_{R}^{r} \rho dV

No. The relevant charge is everything contained within the Gaussian surface. Since the charge only extends to R, we have (for a Gaussian surface with r > R):

Q_{en} = \int_{0}^{R} \rho dV

 

[flyingpig]

But Q_{en} = \int_{0}^{R} \rho dV is for a Gaussian surface that has radius R.

WOuld it be...

Q_{en} = \int_{0}^{R} \rho dV + \int_{R}^{r} \rho dV

That includes inside and outside!?

EDIT: wait, that's no different from my first integral then...

EDIT2: nvm, you are right. My other integral encloses no charge because from R to r, it is just space. I keep thinking my integral is summing field, not charge.

 

[flyingpig]

wait would the integral change if the density was uniform?

 

[flyingpig]

Actually I should show you the final answer first to repay your time you put into helping me

2\pi l\int_{0}^{R} re^{ar} dr = 2\pi l \frac{e^{aR} (aR - 1) + 1}{a^2}

\oint \vec{E} \cdot d\vec{A} = 2\pi l \frac{e^{aR} (aR - 1) + 1}{\epsilon_0 a^2}

\vec{E} = \frac{e^{aR} (aR - 1) + 1}{\epsilon_0ra^2} \hat{r}