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Interesting problem, I almost got it right.

  • Thread starter flyingpig
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  • #1
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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 r1

For part iii), this was also interesting because I also got this part right, but I was a little doubtful on r2. 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 r2, 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 > r2? 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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Answers and Replies

  • #2
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Actually it is possible that u havent understood Gauss's law properly.In cylindrical type of charge distribution we generally take a cylindrical gaussian surface mainly because the electric field is perpendicular to the curved surface of the cylinder and so it helps with the integration of flux.Now the expression on the RHS of the Gauss's law requires the net charge enclosed by the gaussian surface. For r1<r<r2 the gaussian cylinder encloses the entire volume from r=0 to r so the net charge will be contributed by the portion of the nonconducting cylinder inside the surface plus the line charge as it also lies in the region enclosed by the Gaussian surface. Similarly for r>r2 the charge contribution is from the entire non conducting cylinder and the line charge. We are not considering the region upto r in your last question because there is no charge in the region from r=r2 to r and so has no contribution to the net charge enclosed by the gaussian surface.
I hope i made myself clear. In any case first grasp the concept of any topic in physics completely before trying to solve problems. Thank You.
 

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