Electric field inside a conducting cylinder with wire

1. Oct 13, 2010

Liquidxlax

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

You don't really need to read the top paragraph.

A Geiger counter is used to detect ionizing radiation. The detector consists of a thin wire that is surrounded by a concentric circular conducting cylinder. A high voltage is applied to the wire so that it has a positive charge and the surrounding cylinder has the same amount of negative charge. This establishes a very strong electric field inside the cylinder. A low pressure inert gas is inside the cylinder, so that when radiation enters the cylinder it ionizes a few of the gas atoms, and the resulting free electrons are attracted to the positive wire. The e field is so strong as to enable the gas atoms between collisions to gain enough energy to ionize these atoms as well, creating more free electrons, and a chain reaction ensues. An "avalanche" of electrons reaches the wire that is collected by the wire and generates a signal.

Suppose the radius of the central wire is 25x10^(-6)m, the cylinder has a 0.014m radius and the cylinder length is 0.16m. The electric field magnitude at the cylinder wall is 2.9x10^(4) N/C. What is the amount of charge and it's polarity on the central wire?

2. Relevant equations

Er = (4pikQ)/(2pirL) cylinder charge distribution

rho = Q/L

surface area cylinder 2piRL

3. The attempt at a solution

E = kQ/2piRL + KQ/2pirL

can't get the latex or w.e to work hold on

2. Oct 13, 2010

collinsmark

Hello Liquidxlax,
Okay, that equation is good for the electric field in the region outside of a very long, charged wire or cylinder. Which is good, because we are concerned with a region outside of a central wire for now.

If you didn't derive the equation yourself, you can use Gauss' law to derive it, if you wanted to. :

$$\oint_S \vec E \cdot d \vec A = \frac{Q_{enc}}{\epsilon_0}$$

$$E(2 \pi r \ell) = \frac{Q_{enc}}{\epsilon_0}$$

$$E = \frac{Q_{enc}}{(2 \pi r \ell)\epsilon_0}$$

And noting that $k = 1/(4 \pi \epsilon_0)$

$$E = \frac{4 \pi k Q_{enc}}{2 \pi r \ell}$$

[Edit: And by the way, r here is the radius of the hypothetical Guassian surface, not the radius of the wire itself. It is assumed that r is greater than the radius of the wire, since we're trying to find the electric field contribution from the wire, in the region outside of the wire.]

The reason why I went through all of that was to demonstrate Gauss' law. I'd like you to use it again below.
So far so good.
For our purposes, yes, that's right (we can ignore the end-caps).
Umm, :uhh: I'm not sure where that came from either.

It seems to me like you are trying to find the electric field contribution of the charged, hollow cylinder, and incorrectly applying it inside the cylinder. But you're not doing something right.

You might be able to easily figure it out by finding the electric field contribution of a charged, hollow cylinder, inside the cylinder.

You can use Gauss' law to derive that too. Forget about the wire in the center for the moment (we've already found the electric field contribution of that). Use Gauss' law to find the electric field inside of a thin, hollow, charged cylinder.

$$\oint_S \vec E \cdot d \vec A = \frac{Q_{enc}}{\epsilon_0}$$

I'll let you take it from here (hint: the charge enclosed within the region inside of a thin, hollow, charged cylinder is zero -- there isn't any charge on the inside, it's all on the cylinder's surface! ).

Last edited: Oct 13, 2010
3. Oct 13, 2010

Liquidxlax

lol well if that is true, what is the significance of the cylindrical shell around the, but not to signify a radius at which an electric field was measured?

ps. my prof sucks and sucked at trying to teach gauss's law. So my understanding of it isn't very good

oh and i'm not sure about the polarity thing still

4. Oct 13, 2010

collinsmark

For regions inside the thin, hollow, long cylinder, the charge on the cylinder has no bearing on the electric field inside.

This is the same idea for a hollow, spherical shell. The electric field inside of a hollow spherical shell (with uniform charge distribution about its surface) is always zero, as long as there are no other charges around.

So what is the significance of the cylinder at all for this particular problem? Well, ask yourself what is the electric field in the region outside of the cylinder (with the oppositely charged wire inside). Of course, before you get there, your first question is what is the total charge enclosed within the Gaussian surface outside of the cylinder. But now there are two things that are charged: the wire and the cylinder.
Gauss' law is a great tool, but it takes practice.
The problem statement says that the wire has a positive charge, and the cylinder has a negative charge of the same magnitude. In other words, the charge on the wire is Q, and the charge on the cylinder is -Q.

5. Oct 13, 2010

Liquidxlax

yep but he started it yesterday and we have a midterm on tuesday... and 3 assigments due that tuesday and another midterm the next day. So practice is limited

but thanks for the help, i think i'm understanding it better