Homework Help: Electric field question

1. Oct 1, 2008

jaejoon89

1. The problem statement, all variables and given/known data

A pipe that is not a conductor has a uni form charge density 40 C/m^3. The inner radius is 0.15 m and the outer radius is 0.25 m. What is the electric field magnitude at r=0.3 m?

2. Relevant equations

Gauss's Law: net flux = int[E*dA] = Q_encl / epsilon naught

3. The attempt at a solution

First, I figure to treat this as a cylinder, V = pi*r^2 * L

But how would the inner/outer radii come into play? Would it be instead V = pi*r_1^2 * L - pi* r_2^2 * L where r_1 = .25 m (outer radius), r_2 = .15 m (inner radius)? Is that correct so far?

What would the area be? For a cylinder, it's 2*pi*r^2 + 2*pi*r*L, but then L wouldn't totally cancel out, right?

Last edited: Oct 1, 2008
2. Oct 1, 2008

Gear300

You don't really need to worry about the inner and outer radius too much considering they ask you for the electric field outside the pipe. Convert the volume charge density to linear charge density. You can then just treat it as a rod of charge.

3. Oct 1, 2008

jaejoon89

Ok, so is it E = lambda*L (with L canceling) / (epsilon * 2 * pi * r^2) with r = 0.3?
(I must be making a mistake somewhere because i keep getting the wrong answer but this is the only equation that makes a lot of sense)

Last edited: Oct 2, 2008
4. Oct 2, 2008

Phrak

What's relevant, here, is that the charge is distrubuted symmetrically around the axis.

To apply Gauss' Law, you just need to know the enclosed charge per unit length, and the symmetry you can use.

The total charge per unit length is the volume per unit length times the charge density.

5. Oct 2, 2008

jaejoon89

Thanks,

Would the volume per unit length here be
V = pi*r^2 * L where r is the outer radius?

In other words, do you not even have to account for the inner radius here since the electric field is to be calculated outside the cylinder?

Last edited: Oct 2, 2008