So I'm a little confused. My question is: If a conductor is placed in an external electric field, do the field lines penetrate the conductor? My original thought was yes they do and then the induced field inside the conductor cancels out the external field so that the net field inside the conductor is zero. But, in griffiths (3rd edition p. 99), he introduces the problem of a conductor with a cavity inside it. The cavity encloses a single point charge q. Griffiths says "But in a remarkable way the cavity and its contents are electrically isolated from the outside world by the surrounding conductor. No external fields penetrate the conductor; they are canceled at the outer surface by the induced charge there." So is it that in this particular problem an external field wouldn't penetrate the conductor or the cavity? Thanks.
No. (Not in steady-state.) The thing to remember about a conductor is that it contains mobile charges. The external field moves charges around on the surface of the conductor so that the induced field due to the displaced charges cancels out the field that would otherwise have been inside the conductor. Therefore there is zero field inside the conductor - therefore no field lines. If you have a conductor with a cavity - you should distinguish between the space in the cavity - which does have an electric field in it - and the conductor itself - which doesn't. What Griffiths is talking about is the effect of a faraday cage... the field inside the cavity due to the external field is zero.
Your original thought is correct; we can consistently explain the zero (mean) field inside the conductor (in electrostatic conditions) as the sum of the field arising from external charges and the field arising from charges induced on the surface of the conductor. So, with this explanation in mind the external field does penetrate the conductor, but there is also an equal and opposite field. So it is just as if the field arising from external charges didn't penetrate - which is, presumably, Griffiths's point of view in the sentence you quote. As for field lines, these give the directions of the resultant field, not just the field due to the external charges, so naturally they stop at the surface of the conductor, as the resultant field inside the conductor is zero.
Ah it makes sense that the field lines would only represent the net field and not all the individual fields together because it could be messy with 4 individual fields at work! In my mind, I always think of each field having field lines that penetrate everything and in some cases will cancel with other field lines (inside a conductor). With this picture, the field lines from an external field would penetrate through the conductor and into the cavity. But clearly this isn't the right picture. I have one more question: with the same conductor and cavity, but this time without the charge in the cavity, Griffiths argues that the field inside the cavity is zero by use of a closed loop partly inside the cavity and partly inside the conductor. He uses the known relation that the closed line integral of E dot dl is zero and thus the field inside the cavity must be zero. Why can't you use the same logic to argue the field is zero when there is a charge inside the cavity?
Try the same argument for a capacitor ... make the charged plates very thick and put the loop real close to the middle of the plates to avoid edge effects. Part of the loop is in a uniform E field, and part is inside the conductor. Now do the path integral around the loop. I think the situation is easier to see via Gauss' Law.
That calculation tells me that in that configuration the closed line integral of E dot dl ≠ 0 This makes sense to me. But isn't the (closed) integral ∫E dot dl always zero (since the curl of E is always zero?
qE would be the force, do qE.dl would be the work done moving distance dl, and the work done in a closed loop must be zero if E is to be a conservative force. So there is something wrong with your calculation. Redo - pick a square loop. Remember it's a vector product. This should also tell you something about the argument in Griffith. It is easier to use Gauss' Law. The net flux though a closed surface is proportional to the amount of charge enclosed.
Ok it makes sense now! You're right, the calculation correctly done should reveal that the closed integral of E dot dl is zero. Since this integral is path independent, any path will do. The reason he doesn't use the same argument when there is charge present inside the cavity is because all the argument would prove is that the closed integral of E dot dl is zero. It wouldn't prove that the field inside the cavity is zero (which it shouldn't be, anyways). Thank you for your help Simon and Philip.
Well done. I could have just told you this of course - but, this way, you have just learned how to think about puzzles in a way that will help in other situations. Have fun ;)