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 e(DB)^2/2m where D is the separation of the plates and e/m is the specific charge of the electron.
The Attempt at a Solution
Ok well I've found the force due to the electric field created by the capacitor by saying:
F=qE and E=V/D
The charge of the electron is e/m so:
Then you've got the force due to the magnetic field which is trying to pull the electron in a circle. I'm not sure where you go from here. All I know is that:
F=evB but what do I use for v as obviously it's accelerating?