Electric field of a rod

1. Sep 2, 2007

xaer04

1. The problem statement, all variables and given/known data
"An infinitely long rod of radius R carries a uniform volume charge density $\rho$. Show that the electric field strengths outside and inside the rod are given, respectively, by $E=\rho R^2/2\epsilon_0 r$ and $E = \rho r/2\epsilon_0$, where r is the distance from the rod axis."

distance from axis to gaussian surface = r
length of gaussian surface = L
charge per unit volume = $\rho$

2. Relevant equations
surface area of a cylinder (not including bases)
$$A = 2\pi rL$$

Volume of a cylinder
$$V = \pi r^2 L$$

Gauss's Law
$$\Phi = \int \vec{E} \cdot \vec{dA} = \frac{q_{encl}}{\epsilon_0}$$

$$E = \frac{\rho R^2}{2 \epsilon_0 r}$$

$$E = \frac{\rho r}{2 \epsilon_0}$$

3. The attempt at a solution
First off, the surface area doesn't include bases because they're not relevant. I drew electric field lines on the diagram, and the one that is a side view of the rod shows that because it extends off to infinity, the electric field lines won't angle off because there are no ends, making them appear parallel from the side. This means, i don't need the ends of my gaussian cylinder in my surface area - for all practical purposes they're still there (and i'll still need to consider area when dealing with volume charge density), but because they're parallel to the field there is no flux. when viewed from the end (cross section), the lines are radial, which means i dont have to worry about $cos \theta$.

the dotted lines in the cross section show the gaussian surfaces i chose, and the cylinder surrounding the rod shows the outer gaussian surface i chose. the length of that, L, is the same for both surfaces. The outer one i got to work out:

there's no need for an integral, so i have $E \cdot A = q_{encl}/\epsilon_0$
$q_{encl}$ is just charge density times volume.

$$2\pi rLE = \frac{\rho \pi r^2 L}{\epsilon_0}$$
which simplifies to:
$$E = \frac{\rho r}{2 \epsilon_0}$$

when i try to calculate the inner one, i know there is some ratio dealing with object volume and gaussian surface volume, but i don't know where it comes from or how it works... or how to get it:( please help.

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2. Sep 2, 2007

Staff: Mentor

Within the rod you work out the enclosed charge exactly the same way: charge density times volume. What's the volume of your Gaussian cylinder of radius r and length L?

3. Sep 2, 2007

learningphysics

Actually the formula you worked out is the inner one... When you work out the outer one, you'll be using big R to calculate the volume of the charge...

when you do the inner one... r = R... but when you do the outer one... r > R

4. Sep 2, 2007

xaer04

that would be $\pi r^2 L$. i thought it would be that. where i get tripped up is in the example in the text, the same problem is done with a sphere using a ratio of the gaussian inner area and the area of the actual sphere:
$$\frac{4}{3}\pi r^3$$ for the inside area

$$\frac{4}{3}\pi R^3$$ for the sphere area
it states that their areas have a ratio of $r^3/R^3$. it also states that the sphere has a total uniform charge of Q, not a charge distribution based on volume, and they equate $q_{encl}$ to $Q(r^3/R^3)$ for $r<R$.

their final answer looks like this for inside a sphere:
$$E=\frac{Qr}{4\pi \epsilon_0 R^3}$$

if i apply that same logic to my situation, i start by getting the ratio of areas:
$$\frac{\pi r^2 L}{\pi R^2 L}$$

for a ratio of $r^2/R^2$.

my $q_{encl}$ from the first answer will now just be Q, which i will multiply by the area ratio.

$$E(2 \pi r L) = \frac{r^2}{R^2} \left( \frac{\rho \pi r^2 L}{\epsilon_0} \right)$$

which simplifies to:
$$E = \frac{\rho r^3}{2 \epsilon_0 R^2}$$

the r and R are completely wrong.

5. Sep 2, 2007

xaer04

ooh, thanks for spotting that, learningphysics...

6. Sep 2, 2007

xaer04

ooh, and that means my ratio was backwards, which means... it works out, heh. but why do they use the ratio inside of a sphere where i have to use it outside of a rod?

7. Sep 2, 2007

learningphysics

That's the ratio of the volumes... not the areas. surface area of a sphere is $$4{\pi}R^2$$

8. Sep 2, 2007

learningphysics

If they gave you Q instead of $$\rho$$... then you might be get similar results... they won't be exactly analogous... because the charge enclosed in a volume varies as r^3 for the sphere... but only varies as r^2 for the cylinder...

9. Sep 2, 2007

xaer04

alright.. anyways, thanks a lot for the help.