Recent content by jcortez91

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...

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?

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...

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...

Okay thanks. It was bugging me.

In Rohlf's Modern Physics textbook it reads: "The value of q/m for the electron determined by Thomson wan substantially smaller than the values of q/m determined by electrolysis, that is, q/m for the electron is much smaller than for ionized atoms. There were two extreme possibilities: (1) The...