The book also asks for E versus x.
The Attempt at a Solution
This question seemed fairly straightforward: I assumed it could be treated like two +-50 nC capacitors, each with d=1 cm. According to my calculations, (from x=2 cm to x=3 cm for example) E would be 1.4 x 10^7 V/m, and V would be -Es=-1.4 x 10^5 V. The answer key says that E=+- 1.4 x 10^8 V/m, and that V= 140,000 V... I can't figure out how they got E to be 10^8, and if so, what it could have been multiplied by to get 1.4 x 10^5.
I googled it and found someone give an alternate solution in which he/she calculated seperate sigma for the three electrodes (50 nC=Q for the first, 100 nC=Q for the second), and then added them and divided by epsilon0 to get the electric field...I didn't know you could do this; is it still a capacitor then, if the charges are different on either plate..? In any case, this answer was still nothing near the book's, off by a factor of 10^5 or something.