E-B crossed fields parallel plate

[unscientific]

Homework Statement

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.

Homework Equations

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..

 

[TSny]

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

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.

 

[unscientific]

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.

define x as upwards, y into the plate.

ma_x = eV/d

ma_y = qvB

x² + y² = d²

is that right?

 

[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 a_x.

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).

 

[paranormal]

Your attempt at a solution is correct. The equation for the radius of the electron's trajectory is given by:

R = mv/Bq

Where m is the mass of the electron, v is its velocity, B is the strength of the magnetic field, and q is the charge of the electron.

Substituting the given values:

d/2 = m*√(2*e*0.5V/m) / (Bq)

Simplifying and rearranging:

d = 2mv/Bq * √(eV/m)

Since the electron is released from rest at the 0V plate, its initial velocity is 0. Therefore, the equation simplifies to:

d = 0 * √(eV/m) = 0

This means that the electron will not reach the other plate if the distance between the plates (d) is greater than or equal to 0, which is always true.

Therefore, if V is less than e(DB)2/(2m), the electron will not reach the other plate because the distance between the plates will be greater than 0. This is in line with the statement in the problem that the electron will not reach the other plate if the potential difference (V) is less than a certain value.

In conclusion, your solution is correct and it shows that if the potential difference (V) is less than e(DB)2/(2m), the electron will not reach the other plate due to the distance between the plates being greater than 0.