Electromagnetism: Gauss's Law

In summary: I'll let you take it from here. The other two integrals are easy. The answer should be in terms of \rho_0 .Oh wait you are right, it's near my bedtime too...So then for a) E = (ρr^2)/ε ?You're really close. Just remember to use \rho_0 instead of \rho and you should have it.
  • #1
wolski888
25
0

Homework Statement


Consider an infinitely long charged cylinder of radius R, carrying a charge whose density varies with radius as ρ(r) = ρ[itex]_{o}[/itex] r. Derive expressions for the electric field (a) inside the cylinder (i.e. r<R), and (b) outside the cylinder (i.e. r>R).

Homework Equations


Gauss's Law
q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

The Attempt at a Solution


(a) E inside cylinder
I sketched a Gaussian surface inside of the cylinder.
I believe that E is parallel to ds ( [itex]\vec{E}[/itex]||d[itex]\vec{s}[/itex] )
So, gauss's law becomes E[itex]\oint[/itex]ds = q/[itex]\epsilon[/itex] for the side

I believe the integral of ds is 2[itex]\pi[/itex] r L (L being the length of the cylinder even though it is infinite.
And q = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L
derived from q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

So we have E (2[itex]\pi[/itex] r L) = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L /[itex]\epsilon[/itex]
Simplifying to E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ 2[itex]\epsilon[/itex]

Is this correct for (a)?
And for (b) would it be the same idea but with a gaussian surface outside of R?

Thanks!
 
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  • #2
I responded to this in your other thread, but I'll repeat here for completeness.
wolski888 said:

Homework Statement


Consider an infinitely long charged cylinder of radius R, carrying a charge whose density varies with radius as ρ(r) = ρ[itex]_{o}[/itex] r. Derive expressions for the electric field (a) inside the cylinder (i.e. r<R), and (b) outside the cylinder (i.e. r>R).

Homework Equations


Gauss's Law
q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

The Attempt at a Solution


(a) E inside cylinder
I sketched a Gaussian surface inside of the cylinder.
I believe that E is parallel to ds ( [itex]\vec{E}[/itex]||d[itex]\vec{s}[/itex] )
So, gauss's law becomes E[itex]\oint[/itex]ds = q/[itex]\epsilon[/itex] for the side

I believe the integral of ds is 2[itex]\pi[/itex] r L (L being the length of the cylinder even though it is infinite.
And q = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L
derived from q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

So we have E (2[itex]\pi[/itex] r L) = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L /[itex]\epsilon[/itex]
Simplifying to E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ 2[itex]\epsilon[/itex]
So close! :cry: Try the final simplification once more. I think you forgot to cancel something out.
And for (b) would it be the same idea but with a gaussian surface outside of R?
Yes. The trick is to just be careful about determining q. When outside the cylinder, is the total charge q within the Gaussian surface a function of r or a function of R?

=====================
Edit:

I think I see the problem now. Take another look at what I've highlighted in red:
I believe the integral of ds is 2[itex]\pi[/itex] r L (L being the length of the cylinder even though it is infinite.
And q = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L
derived from q=[itex]\rho[/itex] [itex]\delta[/itex][itex]\tau[/itex]

So we have E (2[itex]\pi[/itex] r L) = ρ[itex]_{o}[/itex] r [itex]\pi[/itex] r[itex]^{2}[/itex] L /[itex]\epsilon[/itex]
Simplifying to E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ 2[itex]\epsilon[/itex]
 
Last edited:
  • #3
Thanks for the post!
I don't know what is wrong with the 'r'. Are you saying it should not be there? But, ρ(r) = ρ[itex]_{o}[/itex] r. So I put in the 'r'. Hmm are you saying that [itex]\delta[/itex][itex]\tau[/itex] is equal to: [itex]\pi[/itex] r L?
 
  • #4
Actually, let me give you a better hint than my previous one (admittedly, it probably wasn't a very useful hint).

Evaluate q again. I think you pulled an r out from under the integral sign when you shouldn't have.

dq = ρ dL dr rdθ

Where

ρ = ρ0r

Remember, ρ is a function of r so you can't pull it out from under the integral. You can pull ρ0 out, but not ρ. (i.e., make sure you substitute ρ0r in for ρ *before* you integrate. :wink:
 
Last edited:
  • #5
Where did you get delta from?

For (a)

[tex]Q_{en} = 2 \pi \rho_0 \ell \int_{0}^{R} r^2 dr[/tex]

Do you see why it is from 0 to R?
 
  • #6
I will give it a shot tomorrow (its 12am here...)

flyingpig, who is the question directed to?
 
  • #7
wolski888 said:
I will give it a shot tomorrow (its 12am here...)

flyingpig, who is the question directed to?

Who else but you...?
 
  • #8
flyingpig said:
Where did you get delta from?

For (a)

[tex]Q_{en} = 2 \pi \rho_0 \ell \int_{0}^{R} r^2 dr[/tex]

Do you see why it is from 0 to R?
Actually, that sounds great for part (b). But for part (a), I would write,

[tex]Q_{en} = 2 \pi \rho_0 \ell \int_{0}^{r} r'^2 dr'[/tex]
where [itex] r' [/itex] is a dummy variable.
 
  • #9
collinsmark said:
Actually, that sounds great for part (b). But for part (a), I would write,

[tex]Q_{en} = 2 \pi \rho_0 \ell \int_{0}^{r} r'^2 dr'[/tex]
where [itex] r' [/itex] is a dummy variable.

Oh wait you are right, it's near my bedtime too...
 
  • #10
So then for a) E = (ρr^2)/ε ?
 
  • #11
I don't understand where the 2 is coming from in the integral... is it part of the integration of a cylinder's volume?
 
  • #12
I have E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ε

Where r < R.

Is that right?
 
  • #13
wolski888 said:
I have E = ρ[itex]_{o}[/itex] r[itex]^{2}[/itex]/ε

Where r < R.

Is that right?

Not quite. (See below)

wolski888 said:
I don't understand where the 2 is coming from in the integral... is it part of the integration of a cylinder's volume?

You need to integrate the charge over the volume, meaning there's three dimensions involve. We'll use cylindrical coordinates, since that's the easiest for this particular problem.

One dimension is along the length of the cylinder, along the length of L. Let's call the differential length [itex] d \ell [/itex]

Another dimension is along the radius r. Let's call the differential length of the radius [itex] dr' [/itex] (where [itex] r' [/itex] is a dummy variable, so as not to confuse it with [itex] r [/itex], one of the integration limits [the radius of the Gaussian surface]).

One remaining dimension is the direction perpendicular to [itex] r' [/itex], along [itex] \theta [/itex]. But this is a special differential length, because the length of this one not only varies with [itex] d \theta [/itex], but also with [itex] r' [/itex]. So this differential length is [itex] r' d \theta [/itex].

[tex] dq = \rho \ d \ell \ dr' \ r'd \theta [/tex]

Here, [itex] \rho [/itex] is the charge density, a function of r', and should not be confused with [itex] \rho_0 [/itex]. We haven't made the substitution yet.

Integrating both sides we have,

[tex] q = \int_{r' = 0}^r \int_{\theta = 0}^{2 \pi} \int_{\ell = 0}^L \rho \ d \ell \ dr' \ r' d \theta = 2 \pi L \int_0^r \rho r' dr'[/tex]

Now make your [itex] \rho = \rho_0 r' [/itex] substitution and evaluate the last integral. :wink:
 
  • #14
so it is E = (ρ°r^2)/(3ε°) ?
 

What is Gauss's Law?

Gauss's Law is a fundamental law in physics that relates the distribution of electric charge to the resulting electric field. It states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.

What is the significance of Gauss's Law?

Gauss's Law is significant because it allows us to calculate the electric field for a given charge distribution, and vice versa. This helps in understanding and predicting the behavior of electric fields and charged particles.

How do you apply Gauss's Law in practice?

To apply Gauss's Law, you need to first determine the symmetry of the charge distribution and choose an appropriate Gaussian surface. Then, you can calculate the electric flux through the surface and use Gauss's Law to relate it to the enclosed charge and electric field.

What is the difference between Gauss's Law and Coulomb's Law?

While both laws relate to electric charges and fields, Gauss's Law is more general and can be applied to any type of charge distribution, while Coulomb's Law is limited to point charges. Additionally, Gauss's Law is a statement about the electric field, while Coulomb's Law is an equation for the force between two point charges.

Can Gauss's Law be applied to magnetic fields?

No, Gauss's Law only applies to electric fields. There is a similar law for magnetic fields, called Gauss's Law for Magnetism, but it is not as widely used as Gauss's Law for electric fields.

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