• Support PF! Buy your school textbooks, materials and every day products Here!

Motion of charged particles in E and B fields

  • Thread starter georgia
  • Start date
  • #1
8
0

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?
 

Answers and Replies

  • #2
135
0
Pointing out that specific charge is NOT charge of the electron.
So F/m = eV/mD
F=ma
So accleleration towads the plate by the electric force is = eV/mD

Try relating it with centripetal acceleration caused by the magnetic field.
 
  • #3
960
0

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?
It might be best to use potential energy/kinetic energy relation for a particle that just grazes the plate, eg 1/2mv^2=eV. What else must v be? (use quantums idea about centripetal force from the magnetic field)
 
  • #4
8
0
Ok so at the plate with zero potential the photoelectron has negligible kinetic energy and potential energy of eV. At the other plate the photoelectron has no potential energy and all kinetic energy?

So: 1/2mv^2=eV Is this correct?

The radius of the circle if the photoelectron just grazes the plate will be D so I can say that the centripetal force is mv^2/D and this is equal to the force due to the magnetic field (?)

So mv^2/D=evB
and so v=eBD/m

Substituting into: 1/2mv^2=eV gives:

V=e(DB)^2/2m so have I done it?!

If this is right then thanks everyone for your help :smile:
 
Last edited:
  • #5
960
0
methinks so. But remember this is for particle that never makes it. so for particle that reaches other side, v> that quantity as asked for in statement in problem.
 

Related Threads on Motion of charged particles in E and B fields

Replies
1
Views
433
Replies
12
Views
519
Top