2. This is a schematic of a Geiger Counter. These are the things that go "clickety-click" when radiation is around: you've probably seen someone using one in a movie to detect radiation. I'll pass one around class at some point so you can see how it works (remind me if I forget, since we won't actually cover radioactive decays in the class). How does it work? There's a wire with positive charge running down the middle of a cylinder of an equal negative charge (both conductors). That makes an electric field inside the cylinder, which is filled with a low pressure gas. If a bit of ionizing radiation zips through (see diagram), it whacks loose an electron from the gas. The resulting electron/ion pair are accelerated in opposite directions by the electric field. The electron is accelerated enough that when it runs over another bit of gas, it clobbers some more electrons loose, which do the same thing, snowballing up a pile of electrons headed towards the positive wire. A bunch of moving electrons is a current (we'll do this to death in Ch.25): put that current over a speaker and it goes "click". So: if the radius of the central wire is 25µm, the radius of the shell 1.4 cm, and the cell is 16 cm tall, what's the total positive charge you have to add to make the electric field 2.9x104 N/C just inside the shell wall? 16 cm turns out to be pretty far from infinitely long, but pretend it is for purposes of worrying about which parts of a gaussian surface get flux through them or not. - Particle Signal Charged wire Charged cylindrical shell +44+44

Gauss' law states that the the total electric field flux outside a volume is directly proportional to the charge contained in it. since the dimensions of the cylinder and electric field is given, Gauss' law can be used to determine the charge inside the cylinder to give the mentioned electric field strength. But one has to make sure that the gaussian surface chose should not go beyond the outer shell as, if such surface is considered(i.e. containing inner wire as well as outer shell) the net charge inside the volume covered by this surface is going to be zero and hence electric field strength, so the radius of the gaussian surface should strictly be kept just under the surface to avoid zero field.

The Gauss' law says,

$$\begin{gathered} \oint \overrightarrow{E}·d\overrightarrow{A}=\frac{Q}{{\epsilon }_0} \left(1\right) \\ d\overrightarrow{A} :\text{ Surface area element of the chosen gaussian surface} \end{gathered}$$