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## Homework Statement

Consider the picture below. The picture below shows an infinitely long cylindrical line of charge with charge density per unit length [tex]+\lambda[/tex] and it is located at the center of a non-conducting cylindrical shell with an infinite length and has a charge density per unit volume of [tex] +\rho[/tex]. Using Gauss's Law, calculate the magnitude of the Electric Field as a function of r from the center of cylindrical shell.

i) [tex]r < r_1[/tex]

ii) [tex]r_1 \leq r \leq r_2[/tex]

iii)[tex]r > r_2[/tex]

[PLAIN]http://img840.imageshack.us/img840/620/unledsj.png [Broken]

**Solutions**

[PLAIN]http://img263.imageshack.us/img263/6456/unledctz.png [Broken]

[PLAIN]http://img801.imageshack.us/img801/7536/unledci.png [Broken]

[PLAIN]http://img192.imageshack.us/img192/9096/unledsr.png [Broken]

## The Attempt at a Solution

Alright, I pretty nailed all of them, but I've encountered some interesting problems along the way that I wasn't 100% sure of.

First of all, when they say nonconducting, I immediately assumed it is a insulator, but then I thought, it could also be a semi-conductor, how do you decide which one to go with? Now onto the bigger important question

For i), I pretty much did what they did and got [tex]|\vec{E}| = \frac{\lambda}{2\pi \varepsilon _0 r}[/tex]

For ii), I got only the part of [tex]|\vec{E}| = \frac{\rho(r^2-r_{1}^2)}{2 \varepsilon _0 r}[/tex]

Now here is my question is, why did they even include the E-field from the line? The inequality clearly says [tex]r_1 \leq r \leq r_2[/tex] and the one for the line is [tex]r < r_1[/tex], which means the E-field found in i) cannot hold when we add them together.

Also isn't our Guassian surface only enclosing the Gaussian volume between r (where r is the radius of the Gaussian volume inside the nonconducting cylinder) and r

_{1}

For part iii), this was also interesting because I also got this part right, but I was a little doubtful on r

_{2}. Why? I initially did this

[tex]\sum Q_{en}= \rho(\pi r^2l - \pi r_1 ^2l) + \lambda l[/tex]

Where r is not just r

_{2}, but a larger Gaussian surface

I decided to go with

[tex]\sum Q_{en}= \rho(\pi r_2^2l - \pi r_1 ^2l) + \lambda l[/tex] in the end because that's the charge "enclosed", but I don't understand why it couldn't have been r? If I was at some distance r > r

_{2}? Does the r in the denominator takes care of the decreasing E-field?

I know this post is ridiculously long but please read at least some of it lol

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