Motion of charged particles in E and B fields

In summary, with a parallel-plate capacitor, photoelectrons are emitted with negligible initial velocity, and 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.
  • #1
georgia
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?
 
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  • #2
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
georgia said:

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
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
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.
 

1. What is the relationship between a charged particle's motion and electric and magnetic fields?

Charged particles, such as electrons or protons, will experience a force when placed in an electric or magnetic field. This force will cause the particle to accelerate or change direction, affecting its motion.

2. How do electric and magnetic fields interact with each other to affect a charged particle's motion?

Electric and magnetic fields are often present together and can interact with each other in complex ways to influence the motion of a charged particle. The exact behavior depends on the strength and direction of the fields and the particle's charge and velocity.

3. What is the difference between a uniform and non-uniform electric or magnetic field?

A uniform field has the same strength and direction at all points, while a non-uniform field varies in strength or direction. For a charged particle, this can result in different forces and affect its motion accordingly.

4. How can the motion of a charged particle in an E and B field be described mathematically?

The motion of a charged particle in an E and B field can be described using the Lorentz Force Law, which states that the force on a charged particle is equal to the product of its charge, its velocity, and the combined strength of the electric and magnetic fields at its location.

5. What are some real-world applications of the motion of charged particles in E and B fields?

Understanding the behavior of charged particles in electric and magnetic fields is crucial in many fields, including particle accelerators, plasma physics, and electronics. This knowledge also plays a key role in technologies such as MRI machines, particle detectors, and electric motors.

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