Feynman field transformations

[harpf]

I'm trying to follow Feynman's explanation on page 26-10 of Volume 2 of The Feynman Lectures on Physics. He describes the electric and magnetic fields in a FoR S' moving between the plates of a condenser. Feynman writes that we see a reduced E and an added transverse B in S'. I've attached a copy of the figure in question.

The transformation formula provided in the text is E' = γE, because B = 0 in S. My thinking is that, according to the formula, E' increases. This effect is also consistent with the S length contraction in S'.

I am confused. Thank you.

 

[harpf]

I have attached additional information in the hope that someone will kindly point out the error in my thinking.

 

[PeterDonis]

My thinking is that, according to the formula, E' increases.

You're right, it does. The "reduced" in the text is in error; possibly a typo or a mis-transcription (since the text was originally transcribed from Feynman's actual lectures).

Another way of seeing that E' has to increase is that $E^2 - B^2$ is an invariant; it's the same in all frames. In frame S, B = 0, so the invariant is just $E^2$, the square of the electric field. In frame S', moving relative to the plates, B' is nonzero, so E' must increase to keep the invariant the same.

 

[harpf]

I appreciate your response. Thank you.