Electric Field Strength Inside a Cylindrical Non-conductor

  • Thread starter clementc
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Homework Statement


Hey everyone,
I'm just studying some physics, and came across this question where I don't know why I'm wrong =(
It's from Physics (5th Ed.) by Halliday, Resnick and Krane - Chapter 27, Exercise 22
Positive charge is distributed uniformly throughout a long, nonconducting, cylindrical shell of inner radius [tex]R[/tex] and outer radius [tex]2R[/tex]. At what radial depth beneath the outer surface of the charge distribution is the electric field strength equal to one half the surface value?
Here's a diagram I drew of my interpretation. It's the side on view:

[PLAIN]http://img690.imageshack.us/img690/613/cylinder.png [Broken]

I did it using the shell theorems (HRK says they work for cylinders) and the equation for an electric field for a point charge.


Homework Equations


For this question, all I used was:
[tex]\rho = \frac{q}{V}[/tex]
[tex]E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}[/tex]
And also a bit of shell theory

The Attempt at a Solution


Let the charge in the entire cylindrical shell be [tex]q[/tex].
[tex]\rho=\frac{q}{V}=\frac{q}{lA}[/tex] where [tex]\rho[/tex] is the volume charge density.
At the outermost surface, using the shell theorems, I treated all the charge as being at the centre of the cylinder.
[tex]\therefore E=\frac{1}{4\pi\epsilon_0} \frac{q}{(2R)^2}[/tex]

[tex]=\frac{1}{4\pi\epsilon_0} \frac{\rho l\pi(4R^2-R^2)}{4R^2}[/tex]

[tex]=\frac{3\rho l}{16\epsilon_0}[/tex]

To get where the E field is half of this, I solve for half this strength at a distance [tex]x[/tex] from the centre:
[tex]E=\frac{3\rho l}{32\epsilon_0}[/tex]

[tex]=\frac{1}{4\pi\epsilon_0} \frac{\rho l\pi(x^2-R^2)}{x^2}[/tex]

[tex]\therefore \frac{3}{32}=\frac{x^2-R^2}{4x^2}[/tex]

[tex]12x^2=32(x^2-R^2)[/tex]

[tex]3x^2 = 8x^2 - 8R^2 [/tex]

[tex]5x^2 = 8R^2 [/tex]

[tex]\therefore x = R\sqrt{\frac{8}{5}} [/tex]

[tex]Depth = R(2-\sqrt{\frac{8}{5}}) = 0.735R [/tex]

But the answer was [tex]0.557R[/tex] beneath the surface, and they used Gauss' law. However, I still don't see where I went wrong. I think the shell theorems still hold instead a cylindrical shell, but I'm not sure.

Could anyone please help me? =(
Thank you very much guys
 
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Answers and Replies

  • #2
Doc Al
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I did it using the shell theorems (HRK says they work for cylinders) and the equation for an electric field for a point charge.
What do you mean by 'shell theorems'? Note that this problem has cylindrical symmetry, not spherical symmetry.

Homework Equations


For this question, all I used was:
[tex]\rho = \frac{q}{V} \\
E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}[/tex]
And also a bit of shell theory
:confused:
 
  • #3
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Oh shell theorem
=D
what i meant was the one which like tells you that if you have a uniformly charged spherical/round surface
1. from outside, the electric field is the same as if all the charge were concentrated at a point in the centre
2. the electric field inside the shell is 0

so when im like at point x, i can ignore all the charge outside and the electric field experienced is that which arises only from the charge inside that circle of radius x
 
  • #4
Doc Al
Mentor
45,247
1,594
1. from outside, the electric field is the same as if all the charge were concentrated at a point in the centre
OK, but for a cylinder that means you can treat the charge as if it were concentrated along a line at the center, not a point.
 
  • #5
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OH LOL OF COURSE!!! *bangs head on table*
im such a dummy =P
thank you so much doc Al thank you thank you thank you =D
in that case gauss' law is much nicer heh
 

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