Can Photoelectrons Overcome Electric and Magnetic Fields in a Capacitor?

[Keano16]

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 eD²B²/(2m) where D is the separation of the plates.

Homework Equations

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.

 

[anathema]

Thank you for your post. Your approach is on the right track, but there are a few errors in your calculations.

Firstly, the equation 1/2*m*v^2 = eV is correct, but the expression for v is incorrect. The correct expression for the velocity of an electron in a uniform magnetic field is v = E/B, where E is the electric field between the plates. In this case, the electric field is not equal to V/D, but rather V/(D/2), because the plates are parallel and the potential difference is shared between the two plates.

Secondly, when substituting v = E/B into the kinetic energy equation, you need to use the correct expression for E as mentioned above. So the equation becomes 1/2*m*(V/(D/2))^2 = eV.

Finally, to get the minimum potential difference required for the electron to reach the positive plate, you need to set the kinetic energy equal to the work done by the electric field, which is eV. So the final equation becomes 1/2*m*(V/(D/2))^2 = eV, which can be rearranged to V = eD^2B^2/(2m).

I hope this helps. Let me know if you have any further questions.[Your Title/Position]