1. The problem statement, all variables and given/known data Photoelectrons are emitted with negligible initial velocity from one plate of a parallel-plate capacitor which is at zero potential. The other plate is held at a potential +V and the capacitor is in a uniform magnetic field of flux density B parallel to the plane of the plates. Show that no electron can reach the positive plate if V is less than eD2B2/(2m) where D is the separation of the plates. 2. Relevant equations 3. The attempt at a solution What I have tried to do is find the minimum amount of energy required to go from one plate to another. Equating the kinetic energy required to overcome this E-field between the plates, I have done: 1/2*m*v2=eV Substituting v=E/B (E and Bi fields are perpendicular), and E=V/D I get V=2eD2B2/m Which means I am a factor 4 higher than the required answer. What am I doing wrong? Thanks.