# Electric Field Strength Inside a Cylindrical Non-conductor

## 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 $$R$$ and outer radius $$2R$$. 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:
$$\rho = \frac{q}{V}$$
$$E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$
And also a bit of shell theory

## The Attempt at a Solution

Let the charge in the entire cylindrical shell be $$q$$.
$$\rho=\frac{q}{V}=\frac{q}{lA}$$ where $$\rho$$ 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.
$$\therefore E=\frac{1}{4\pi\epsilon_0} \frac{q}{(2R)^2}$$

$$=\frac{1}{4\pi\epsilon_0} \frac{\rho l\pi(4R^2-R^2)}{4R^2}$$

$$=\frac{3\rho l}{16\epsilon_0}$$

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

$$=\frac{1}{4\pi\epsilon_0} \frac{\rho l\pi(x^2-R^2)}{x^2}$$

$$\therefore \frac{3}{32}=\frac{x^2-R^2}{4x^2}$$

$$12x^2=32(x^2-R^2)$$

$$3x^2 = 8x^2 - 8R^2$$

$$5x^2 = 8R^2$$

$$\therefore x = R\sqrt{\frac{8}{5}}$$

$$Depth = R(2-\sqrt{\frac{8}{5}}) = 0.735R$$

But the answer was $$0.557R$$ 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.

Thank you very much guys

Last edited by a moderator:

Doc Al
Mentor
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:
$$\rho = \frac{q}{V} \\ E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$
And also a bit of shell theory 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

Doc Al
Mentor
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.

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