1. The problem statement, all variables and given/known data A problem of practical interest is to make a beam of electrons turn a 90∘ corner. This can be done with the parallel-plate capacitor shown in the figure (Figure 1) . An electron with kinetic energy 2.0×10−17 J enters through a small hole in the bottom plate of the capacitor. What strength electric field is needed if the electron is to emerge from an exit hole 1.0 cm away from the entrance hole, traveling at right angles to its original direction? 2. Relevant equations Vf2=Vi2+2ad F=ma F=Eq K (kinetic energy) = 1/2mv2 3. The attempt at a solution The electric field needs bring vertical velocity to zero and horizontal velocity to what the vertical velocity was. I should just be able to find what E-field is necessary to bring the vertical velocity to zero without having to worry about the horizontal velocity, I think. Vi=(2K/m)1/2 d=.01*cos45 0=(2K/m)+2a*.01*cos45 -K/(m(.01*cos45))=a Now finding acceleration in terms of E qE=ma qE/m=a Combining them: -K/(m(.01*cos45))=qE/m -K/(q(.01*cos45))=E Now, when I go through this I get 17655 N/C, which seems close (right order of magnitude), but according to Mastering it's not correct. Any help would be appreciated, thank you!