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Electric field and Guass' Law question

  • Thread starter l1fesavers
  • Start date
  • #1
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Hi, I'm trying to complete the weekly homework questions that have been assigned to me but I am quite stuck. I've searched all over the web and watched a few youtube lectures but I can't seem to wrap my head around Gauss' law. I was hoping someone could help point me in the right direction!

Homework Statement


Imagine an infinite line of charge with charger per unit (+)lamda. Surrounding the line of charge is a cylinder with inner radius a and outer radius b. The cylinder has a charge density of (ALPHA*r). Answer in terms of quantities given or a subset, use Gauss' law.

Units of Alpha?
Electric field for r < a
Electric field for a < r < b
Electric field for r > b

I have seen so many examples of a typical problem with a gaussian sphere around a point charge, but I cannot wrap my head around how to go from that to solving a problem like this. I don't want to solution, just someone to point me in the right direction so I can go from there and hopefully arrive at the correct solution!

Thanks!
 

Answers and Replies

  • #2
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Probably the best way to start would be to calculate the field from the line of charge alone, without worrying about the other cylinder of charge outside of it. Are you familiar with how to do that problem?
 
  • #3
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My instinct is to use Coloumb's law...

E = kQ/r^2

Since the line is infinitely long, I can basically treat it as a point charge?

EDIT

Okay, so that's completely wrong. I'm going to type this out in words (hope it will help me solidify my understanding.

Using the definition of Electric Flux as "Electric field * Area perpendicular to the field", and then also using Gauss' law which says the flux is the enclosed charge over epsilon not, i can come to this:

EA = Q/epsilonNot

Then I know that Q = density*Length, and that the area of my "gaussian surface" is 2pi*r*L.

E * 2pi * r * L = density*L/epsilonNot

Then get the Electric field E:

E = density/2pi*r*epsilonNot

...
 
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  • #4
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Hi l1fesavers,

No you cannot treat an infinite line charge as a point charge.

Let's look at the first question: Finding the units of [itex]\alpha[/itex].

If I place some charge on a hollow cylinder, what kind of charge density do I have? Is it Coulomb's per unit length? Coulombs per unit area? Coulombs per unit volume?
 
  • #5
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Not exactly. This problem is fundamentally different from a point charge problem, because the field doesn't fall off in all three dimensions, just two.

Gauss's Law can tell you the answer to this. For your Gaussian surface, try a cylinder of length L and radius r, surrounding the line of charge. Start by working out how much charge is inside of the cylinder, and then see what you can deduce from that using Gauss's Law.
 
  • #6
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EA = Q/epsilonNot

Then I know that Q = density*Length, and that the area of my "gaussian surface" is 2pi*r*L.

E * 2pi * r * L = density*L/epsilonNot

Then get the Electric field E:

E = density/2pi*r*epsilonNot

...
Bingo. Now try and apply similar reasoning to the case where your surface encompasses the outer cylinder as well.
 
  • #7
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Thanks for the help so far! I just edited my post but I'll put it here as well...

Using the definition of Electric Flux as "Electric field * Area perpendicular to the field", and then also using Gauss' law which says the flux is the enclosed charge over epsilon not, i can come to this:

EA = Q/epsilonNot

Then I know that Q = density*Length, and that the area of my "gaussian surface" is 2pi*r*L.

E * 2pi * r * L = density*L/epsilonNot

Then get the Electric field E:

E = density/2pi*r*epsilonNot

------------------------------------------------------------

As for placing a charge on a hollow cylinder, it would be... Coloumbs per unit area? Because it is on the "surface"... ?
 
  • #8
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As for placing a charge on a hollow cylinder, it would be... Coloumbs per unit area? Because it is on the "surface"... ?
My apologies, I misread the problem. Yes, that would have been correct. But it sounds like we're dealing with a cylinder of some volume, i.e. a cylinder that has had part of its middle cut out (perhaps Chopin can confirm that).

So if we're dealing with a volume, the charge density has what units then?
 
  • #9
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As for placing a charge on a hollow cylinder, it would be... Coloumbs per unit area? Because it is on the "surface"... ?
If I read your problem right, the cylinder has finite thickness, right? If so, you need to be thinking about volume, not area.
 
  • #10
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Yes. The surrounding cylinder has an inner radius a and outer radius b, and has a charge density of alpha*r

So then we're dealing charge per unit volume
 
  • #11
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Yes. The surrounding cylinder has an inner radius a and outer radius b, and has a charge density of alpha*r

So then we're dealing charge per unit volume
Right. So if the units of charge density are charge/volume, what are the units of alpha? Use dimensional analysis.
 
  • #12
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It feels intuitively wrong but this is what I can come up with.

If the density is Q/m^3 and we have it defined as alpha*r, and r is "1 dimension" then for the density to work out to Q/m^3, alpha would have to be..

Q/m^4 ?
 
  • #13
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It feels intuitively wrong but this is what I can come up with.

If the density is Q/m^3 and we have it defined as alpha*r, and r is "1 dimension" then for the density to work out to Q/m^3, alpha would have to be..

Q/m^4 ?
That's correct. But you would want to say "Coulombs per meters^4" or "charge per unit length^4," since Q is not a unit.

Now you've already found the E-field for r < a, let's call it EI to denote the E-field for region I. Then:

[itex]\vec{E}_1 = \frac{\lambda}{2\pi\epsilon_0}\hat{r}[/itex]

Again, you're going to use Gauss' Law to find the E-field for a < r < b, let's call it region II. So we're looking for EII.

First, it may be helpful to completely forget about the line charge, and simply find the field inside of the cylinder. Remember that we're looking for the field as a function of r.
 
  • #14
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That's correct. But you would want to say "Coulombs per meters^4" or "charge per unit length^4," since Q is not a unit.
Are you sure that's right? I'm pretty sure you can't incorporate the radius into the units of the density like that. It so happens that the density is linearly dependent on the radius here, but it wouldn't have to be. What if the density were [itex]\alpha \sin(2\pi r)[/itex], or something screwy like that? If your reasoning applies, what would the units be for that?

I'm pretty sure the units on the density need to just be charge per meters^3, because in order to turn the density into an amount of charge, you need to multiply by a volume, which is meters^3. However, since the density is variable across space, you will also need to integrate over the volume in question--you can't simply do it by straight multiplication like you could if the density was constant.
 
  • #15
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Are you sure that's right? I'm pretty sure you can't incorporate the radius into the units of the density like that. It so happens that the density is linearly dependent on the radius here, but it wouldn't have to be. What if the density were [itex]\alpha \sin(2\pi r)[/itex], or something screwy like that? If your reasoning applies, what would the units be for that?

I'm pretty sure the units on the density need to just be charge per meters^3, because in order to turn the density into an amount of charge, you need to multiply by a volume, which is meters^3. However, since the density is variable across space, you will also need to integrate over the volume in question--you can't simply do it by straight multiplication like you could if the density was constant.
You guys have lost me...
 
  • #16
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would the electric field within the cylinder (inner radius a, outer radius b) just be the electric field @ b minus the electric field @ a?

Also, how do you guys get the formatted equations?
 
  • #17
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Sorry, that was a bit of a tangent. The fundamental thing you need to focus on now is, how do you determine how much charge is inside of a given volume? That's what you need for Gauss's Law to work. With the line of charge it was easy--just multiply the density times the length. For this, it's a bit trickier, because it's a density in three dimensions, and it's also variable across space--namely, it gets higher the closer you get to the center.

You might start by pretending the density is constant, i.e. it's just [itex]\alpha[/itex], instead of [itex]\alpha r[/itex]. Try to work out how much charge is in a cylinder of a given size in that case. Then, generalize to a variable density.
 
  • #18
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would the electric field within the cylinder (inner radius a, outer radius b) just be the electric field @ b minus the electric field @ a?

Also, how do you guys get the formatted equations?
No, it's a bit more complicated than that. Once again, try to think through it step-by-step with Gauss's Law...don't try to take any shortcuts just yet.

To get the formatted equations, use the [ tex ] tag (or [ itex ] tag for inline equations). Those tags use the [itex]\TeX[/itex] math formatting language, which is useful to know for lots of reasons if you're in the math/physics business.
 
  • #19
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I'm pretty sure the units on the density need to just be charge per meters^3, because in order to turn the density into an amount of charge, you need to multiply by a volume, which is meters^3.
That's the reasoning I tried to follow:

[itex][\rho] = \frac{[charge]}{[length]^{3}} = [\alpha] \cdot [length] \rightarrow [\alpha] = \frac{[charge]}{[length]^{4}}[/itex]

Are you're saying the units should be:

[itex][\alpha] = \frac{[charge]}{[length]^{3}}\frac{1}{r}[/itex]

And r should be kept separate?

I double-checked my units for all the E-fields, they appear to work out correctly.
 
  • #20
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The units will happen to work out, but they're lying to you. You have to approach it differently.

Imagine taking a very small box of volume V, so small that the density variance from one end of the box to the other is so small that it's negligible. In that case, the total charge contained within the box is [itex]Q = \rho(r)\cdot V[/itex]. Now imagine the next box over. It's a slightly different amount of charge, because the density is different: [itex]Q' = \rho(r')\cdot V[/itex]. And so on and so on...each box is going to have a different amount of charge in it. To get the total charge, you have to add up all of the little boxes. Also, our assumption that the charge density is constant is not actually true for any finite-sized box. So we also need to take the limit as [itex]V\rightarrow 0[/itex].

You should be starting to get itchy memories of Calculus I at this point--this is the definition of an Riemann sum, also known as an integral. Therefore, what you really need to do is to perform a volume integral of [itex]\rho(r)[/itex] over the cylinder--that's the way to get the total amount of charge for a continuously-varying density. This method will work for any charge density, not just one which is of the form [itex]\alpha r[/itex]--you could have something crazy like [itex]\rho(r) = r^{17} + e^{2 \cos(r)}[/itex] and it would still work, assuming you were clever enough to do the integral.
 
  • #21
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The units will happen to work out, but they're lying to you. You have to approach it differently.
Those pesky units! :p

Imagine taking a very small box of volume V, so small that the density variance from one end of the box to the other is so small that it's negligible. In that case, the total charge contained within the box is [itex]Q = \rho(r)\cdot V[/itex]. Now imagine the next box over. It's a slightly different amount of charge, because the density is different: [itex]Q' = \rho(r')\cdot V[/itex]. And so on and so on...each box is going to have a different amount of charge in it. To get the total charge, you have to add up all of the little boxes. Also, since our assumption that the charge is constant is not actually true for any finite-sized box, we need to take the limit as [itex]V\rightarrow 0[/itex].

You should start to be getting itchy memories of Calculus I at this point--this is the definition of an Riemann sum, also known as an integral. Therefore, what you really need to do is to perform a volume integral of [itex]\rho(r)[/itex] over the cylinder--that's the way to get the total amount of charge for a continuously-varying density. This method will work for any charge density, not just one which is of the form [itex]\alpha r[/itex]--you could have something crazy like [itex]\rho(r) = r^{17} + e^{2 \cos(r)}[/itex] and it would still work, assuming you were clever enough to do the integral.
That makes perfect sense, but that sounds a lot like the Coulomb integral, where you integrate over a volume. But when you're using Gauss' Law you're integrating over a surface (in this case the surface of a cylinder). Even if the charge density ρ varies over the volume of the cylinder, Gauss' Law still works. I'm still not getting how alpha wouldn't have units of charge/length^4.
 
  • #22
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That makes perfect sense, but that sounds a lot like the Coulomb integral, where you integrate over a volume. But when you're using Gauss' Law you're integrating over a surface (in this case the surface of a cylinder). Even if the charge density ρ varies over the volume of the cylinder, Gauss' Law still works. I'm still not getting how alpha wouldn't have units of charge/length^4.
Can you clarify what you mean by "Coulomb Integral"?

Remember that Gauss's Law works off of the "total contained charge", which means it involves both a volume integral and a surface integral. Technically speaking, it's written as:

[tex]\int_V d^3x \frac{\rho(x)}{\epsilon_0} = \int_S d^2S \cdot E(S)[/tex]

Which means you can take the volume integral of the density, and it will be equal to the surface integral of the flux (technically speaking, this is an application of the Divergence Theorem of vector calculus). Often, however, the "volume integral" is so trivial that we don't even think about it as an integral--for instance, if it's a point charge, then we know the total charge right away. In the case of a line of charge, the total charge is the density times the length, as we calculated up above. If it were a constant density, then the total charge is the density times the volume.

All of those calculations are technically volume integrals, but since they're constant, you can perform the integral trivially by just multiplying. In the case of a variable density like this, though, you aren't so lucky. You instead have to perform an integral over the volume in order to get the total charge.
 
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  • #23
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sometimes the correct response is - thank you...
 
  • #24
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By the Coulomb integral, I just mean:

[itex]\vec{E}(\vec{r}) = \int_{V} \frac{1}{4\pi\epsilon_0}\frac{\rho(\vec{r'})}{| \vec{r}-\vec{r'}|^2}\hat{r}[/itex]

Which accounts for a charge density that varies over space.

Let me confirm that I am assuming that alpha is a constant, but even if it were not, the units should still be charge/length^4. You spoke of integrating the charge density to get the total charge:

[itex]Q = \int_{V} \rho(\vec{r'}) dV = \int \int \int \alpha \cdot r \cdot r \cdot dr d\phi dz = \frac{\alpha}{3}[b^3 - a^3] \cdot 2\pi l[/itex]

Where l is the length of the cylinder. Looking at the units, we have:

[itex][Q] = [\alpha][length]^3[length] = [charge] \rightarrow [\alpha] = \frac{[charge]}{[length]^4}[/itex]

Even if alpha was somehow dependent on the position in space and we couldn't pull it out of the integral, it should still work out to have the same units. It wasn't indicated in the problem that alpha does (or might) depend on position.

Anyway, I'm starting to get worried that we're frightening the OP, haha. Have you made any progress l1fesaver?
 
  • #25
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Unfortunately at the moment I am quite stuck
 

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