Electric field and Guass' Law question

[l1fesavers]

Okay so in our case $\rho (r) = \alpha r$. So...

I would integrate, from r=a to r=b of $\alpha r \pi L r^2$

 

[Chopin]

Not quite--think about this integral as being a bunch of cylindrical shells that have an infinitesimal thickness of $dr$. The volume of such a shell is $2\pi rL\:dr$ (imagine unrolling it).

 

[l1fesavers]

I'm heading in this direction... $dQ = \alpha r 2 \pi L r dr$

 

[Chopin]

I'm heading in this direction... $dQ = \alpha r 2 \pi L r dr$

That's it. Now use that to find the total charge.

 

[l1fesavers]

$\int_{a}^{b} \alpha r^2 2 \pi L dr$

$2 \pi L \alpha \int_{a}^{b} r^2 dr$

$2/3 \pi L \alpha (b^3 - a^3)$

 

[Chopin]

Looks right to me. You're in the home stretch now..."finish him"!

 

[l1fesavers]

So now it's pretty much the same as before...

$EA = Q_{enc}/\epsilon_{0}$

$E 2 \pi rL = 2\pi/3 L \alpha (b^3-a^3)$

$E = \dfrac{ \alpha (b^3 - a^3)}{3r\epsilon_{0}}$

edit: really liking this Latex thing, sorry if it's a bit messy still

 

[Chopin]

And there you go. Now you've computed the field in Region I, and for Region III (but don't forget to add the line of charge back in--it contributes to $Q_{enc}$ for all regions.) Last stop is calculating Region II, which you should be able to do from your existing work. As a final check, try confirming that the field strength is continuous as you move between regions.

 

[l1fesavers]

Okay, so I must have conceptually lost something. I was under the impression we were doing region II (a < r < b)?

 

[Chopin]

No, that expression doesn't work for Region II. Remember that you only count the charge which is inside the Gaussian surface, so your volume integral only goes up to $r$. If $r < b$, then the bounds of your integral aren't right.

It might be clearer if you change the name of your integration variable...try calling it $R$ instead of $r$, so that you can keep track of which ones are your integration variable, and which are the radius of your Gaussian cylinder.

 

[l1fesavers]

Woosh. Okay let me start from scratch real quick then. By the way, thank you so much for this walk through, I cannot even begin to express how much it helps and how thankful I am!

This is what I kind of figured...

I was integrating from r = a to b (the "thickness" of the cylinder shell) so that we would find the charge contained within that volume. Why doesn't that correspond to a < r < b?

 

[Chopin]

Because what you're ultimately trying to find is the electric field for any given radius $r$. Gauss's Law says that you can find this by finding the flux on a cylinder of radius $r$, which in turn is proportional to the charge contained inside of it. So what you're trying to work out is the total charge enclosed within a cylinder of radius $r$.

So if $a < r < b$, then we don't actually care about the total charge in the cylinder, only the amount of that charge which falls inside radius $r$. So you need to integrate from $a$ to $r$ only...the remaining charge outside of that radius makes absolutely no contribution to the electric field at $r$. This is a general result, and is, in my opinion, the single most surprising consequence of Gauss's Law.

 

[l1fesavers]

Okay that actually makes good sense to me. So in that case for region II our result would only need to be modified slightly...

$E = \dfrac{ \alpha (r^3 - a^3)}{3r\epsilon_{0}}$

Now you mentioned that I still need to include the original line charge so that would be something like:

$Q_{enc} = Q_{1} + Q_{2}$ where Q1 was line charge and Q2 was the cylindrical shell?

I feel like I'm not doing that right...

 

[l1fesavers]

Going through the process for Region III again but this time including the line charge netted me this:

$$E = \dfrac{\alpha(b^3-a^3)+\lambda}{3 \epsilon_{o} r}$$

 

[E_M_C]

Now you mentioned that I still need to include the original line charge so that would be something like:

$Q_{enc} = Q_{1} + Q_{2}$ where Q1 was line charge and Q2 was the cylindrical shell?

Something like that, but you've already calculated all the terms that you need, you just have to put them together correctly.

You may remember the concept of superposition when dealing with point charges. How would you use superposition to write the total electric field in region II? In region III?

 

[E_M_C]

Going through the process for Region III again but this time including the line charge netted me this:

$$E = \dfrac{\alpha(b^3-a^3)+\lambda}{3 \epsilon_{o} r}$$

Not quite. Was λ divided by 3ε₀r in region I?

 

[l1fesavers]

That's why I suspected I was doing it wrong.

Well the superposition of point charges just says that it's the sum of all the electric fields... so I can really only conclude that its:

$$E_{t} = E_{1} + E_{2}$$ where $$E_{1} = \dfrac{\alpha(b^3-a^3)}{3 \epsilon_{o} r}$$ and $$E_{2} = \dfrac{\lambda}{2\pi r \epsilon_{o}}$$

 

[E_M_C]

Ok. Now which terms do you add together and for what regions?

You're going to have 3 separate fields, one for each region (I, II, and III). What do you get for each region?

 

[l1fesavers]

A little confused by the question...

For field r < a

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}}$$

For field a < r < b

$$E = \dfrac{\alpha(r^3-a^3)}{3 \epsilon_{o} r}$$

For field r > b

The sum of the above two?

 

[E_M_C]

Region II is missing a term. Think about the total charge enclosed. What fields are enclosed when you place a Gaussian cylinder at a < r < b ?

Region III is not the sum of the two you wrote, but close. First, what would the field be in region III, that is r > b, if we only considered the cylinder of charge? Then, what is the field in region III when we consider both the cylinder and the line of charge?

Again, for regions II and III, think superposition.

 

[l1fesavers]

Maybe I'm misunderstanding superposition? That the net result of several fields is the summation of the result of each individual field?

As for region 2, the field that is enclosed is the volume of the hollowed out cylindrical shell?

 

[E_M_C]

That the net result of several fields is the summation of the result of each individual field?

Right.

As for region 2, the field that is enclosed is the volume of the hollowed out cylindrical shell?

Yes, and what else?

 

[l1fesavers]

Well, the only other thing is the line of charge haha

 

[l1fesavers]

So for Region 2

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}} + \dfrac{\alpha(b^3-a^3)}{3 \epsilon_{o} r}$$

 

[E_M_C]

So for Region 2

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}} + \dfrac{\alpha(b^3-a^3)}{3 \epsilon_{o} r}$$

So close! Remember that the Gaussian cylinder for region II has a radius somewhere between a and b.

The term (b³ - a³) implies that we're calculating the field for the entire cylinder. Does region II include the entire cylinder? What might you change about your equation to make the dimensions match that of the Gaussian cylinder (of radius r) placed in region II?

 

[l1fesavers]

Region II does not include the entire cylinder - it only includes up until our Gaussian surface, so it goes to 'r'...

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}} + \dfrac{\alpha(r^3-a^3)}{3 \epsilon_{o} r}$$

 

[E_M_C]

Region II does not include the entire cylinder - it only includes up until our Gaussian surface, so it goes to 'r'...

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}} + \dfrac{\alpha(r^3-a^3)}{3 \epsilon_{o} r}$$

Perfect!

Now you can write the field in region III. Does that region include the entire cylinder?

 

[l1fesavers]

It does, so...

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}} + \dfrac{\alpha(b^3-a^3)}{3 \epsilon_{o} r}$$

 

[E_M_C]

It does, so...

$$E = \dfrac{\lambda}{2\pi r \epsilon_{o}} + \dfrac{\alpha(b^3-a^3)}{3 \epsilon_{o} r}$$

You got it, you're all done. Chopin would be proud!

 

[l1fesavers]

Thanks so much to both of you for your amazing help! And even more amazing level of patience with both my ignorance and extremely rusty math haha