Recent content by 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

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...

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

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}

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

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

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?

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?

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} =...

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}

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...

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...

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

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