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## Homework Statement

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.

## Homework Equations

F=q(vxB)

E=V/D

F=qE

## 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

so F=qV/D

The charge of the electron is e/m so:

F=eV/mD

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?