Electric field and Guass' Law question

[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

 

[E_M_C]

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

It's our pleasure, and you're welcome :)

You did just fine. And as always, practice makes perfect!

 

[Chopin]

Hey guys, sorry I missed the grand finale here. Well done, l1fesavers!

Looking back over the last few posts, my only suggestion would be that you can do the whole problem using only Gauss's Law, without resorting to the superposition principle for the fields. All you have to do is think about the total charge enclosed in each area.

Region I contains only the line of charge
Region II contains the line of charge, plus part of the cylinder (out to radius r)
Region III contains the line of charge, plus the entire cylinder

In all cases, you take this total charge, and divide it by the area of the cylinder to get the field strength. Even in case II and III, where the charge is in two different pieces, it works. The superposition principle of the fields can be viewed of as a consequence of this fact.

 

[Chopin]

Also, E M C, now that I look at this again in the cold light of day, you're right about the units on \alpha. It's a quantity which let's you convert a distance into a density, so its units must be \frac{density}{distance} = \frac{\frac{charge}{m^3}}{m} = \frac{charge}{m^4}. I'm not sure why that wasn't making sense to me last night. That idea still wouldn't work for a case like \rho(r) = \alpha e^r, but I guess units only ever work when you're dealing with linear quantities, so that's not a big deal.

Sorry for the confusion!

 

[l1fesavers]

Okay so using "strictly" gauss' law I believe will look something like this:
(This is region two, a<r<b)

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

Writing out the Latex is daunting, but just substituting this Q_{enc} into the electric field equation should be correct?

 

[Chopin]

Yup, that should do it. You should find once you do the divisions, that you end up with the same result that you would have if you had gone all the way through the field calculations for each blob of charge separately, and then added the fields after the fact.

 

[l1fesavers]

Actually, I just wrote out the entire equation and it simplified right back to the superposition we were doing earlier, so that would confirm that it is the correct answer! ...i think

--edit--
Perfect! Again, thanks for the help.. and i'll be back hah