# E-B crossed fields parallel plate

1. Mar 4, 2013

### unscientific

1. The problem statement, all variables and given/known data

A parallel plate has p.d. V with one end 0V and the other +V. B-field of strength B has direction along plates. An electron is released from rest at 0V plate. Show that if V is less than e(DB)2/(2m) the electron will not reach the other plate.

2. Relevant equations

3. The attempt at a solution

At half-way inbetween the plates, the radius must be d/2

R = mv/Bq

d/2 = m*√(2*e*0.5V/m) / (Bq)

I end up with e(DB)2/(4m) instead..

2. Mar 4, 2013

### TSny

Hello.
I don't follow your statement. Could you just as well reason that when the electron is 3/4 of the way across, then the radius must be d/4? If so, would you get the same answer?

I think you're going to need to set up the equations of motion for the x and y components of acceleration.

3. Mar 5, 2013

### unscientific

define x as upwards, y into the plate.

max = eV/d

may = qvB

x2 + y2 = d2

is that right?

4. Mar 5, 2013

### TSny

Let me make sure I understand your choice of directions and signs. Correct me if I'm wrong. You're choosing the electric field in the negative x direction so that the E field accelerates the electron in the positive x direction. The x-axis is oriented perpendicular to the plates.

y-axis is parallel to the plates and B is parallel to the plates but perpendicular to both the x axis and the y axis such that when the electron starts moving the B field begins to deflect the electron toward positive y.

But once the electron is deflected some in the y-direction by the B field, the electron will then have a y-component of velocity. So, the magnetic field will then cause some force parallel to the x-axis! So, you need to include a magnetic force term in your equation for ax.

The electron is going to move along a curved path, so you can't say x2+y2 = d2 (if d represents the separation of the plates).

Last edited: Mar 5, 2013