# Trouble w/ electron in a television picture tube. Hints please

1. Jan 25, 2010

### kusiobache

1. The problem statement, all variables and given/known data
An electron is accelerated horizontally from rest in a television picture tube by a
potential difference of 25,000 V. It then passes between two horizontal plates 6.5 cm
long and 1.3 cm apart that have a potential difference of 250 V, as shown below. At
what angle θ will the electron be traveling after it passes between the plates?

2. Relevant equations
E=F/Q E=KQ/r^2 F=EQ V=KQ/R E=V/D F=MA

3. The attempt at a solution
I haven't really attempted much, I kind of need a hint to get me started.

I mean, I THINK I know how to solve it but I'm stuck. figure it is accelerated horizontally which would be a force in the X direction, and between the plates it experiences a force in the Y Direction, so you would just do Tan$$\alpha$$ = Fy/Fx

My Thing is, if this is correct, I have no clue how to GET the force X. I mean, you can find the velocity, idk how I could find acceleration from there for F=MA. Or, you could say Fx=EQ, but then I have no clue how to find an electric field at the beginning.

I'm Also going to try doing the arctan function w/ the velocities, however that will work I shall see. -> Just tried it, I used electron volts to get the P.E which backfired

I don't need it solved for me, I just need a hint or two to get me started.

P.S. Just found this, might help (like on page 6 of google) https://www.physicsforums.com/showthread.php?t=13031. I'll try it

EDIT: Just Re-did the electron volt thing for velocities, and I got .57 degrees. Doesn't seem right but meh, I'll check it tomorrow. If anyone gets any ideas though I'm open to advice :)

Last edited: Jan 25, 2010
2. Jan 25, 2010

### Senjai

If its asking for the angle of deflection. First.. Ask yourself.. Is their any acceleration in the vertical direction? (assume the electron is being deflected vertically.) If you have an electron go through the plates, there must be some force exerted on it to change its path of motion. When it leaves the plates it will have been deflected a certain distance from its original position. You can use that distance, and the distance of the plates to find the angle.

positive (+)
__________________
|
| dy
(theta) |
---------------------|
dx

____________________
Negative (-)
Not sure if that diagram helps, but you know dx, the distance or length of the plates. you need to find dy, the deflection along the y axis.

Then you can use trigonometry to determine the angle of deflection, which will remain constant even after it leaves the plates, but look at only the moment it leaves the plates and stops being accelerated.

Using kinematics, we can say dy = deflection..

So..

$$deflection = V_{0y}t + \frac{1}{2}at^2$$

So because their is no initial velocity in the y direction, we're left with

$$deflection = \frac{1}{2}at^2$$

Do you understand the concept?

Now you'll need to make appropriate substitutions for a, and t, using the information the question has provided to derive a usable formula to calculate the deflection, dy. Once you have this, use trig with dx (known) and dy..

If you need any help, ask ill be on for a while. Wont derive it for you :P Make a good effort.

3. Jan 25, 2010

### Senjai

*Hint* What is the net force on the electron when it is passing through the plates, weight is negligible.

EDIT: My theta ended up being 0.72 Degrees, if you have an answer key, let me know if this is correct..

Last edited: Jan 25, 2010
4. Jan 26, 2010

### kusiobache

Ahh I see, it makes sense now. I can use electric fields to get the force in the Y direction. Then, I figure I can find the x velocity somehow (probably w/ the given 25000 voltage, I'll try it), and then use that for time and so on. I'll try it and get back to you later.

And sorry, but I won't get the answer until Thursday (bcz the teacher hasn't done it yet apparently), but I'll let you know what I get tonight and I'll let you know what the teacher gets once he does it. (I figure that he HAS done it, but lost whatever sheet he had the work and answer on).

5. Jan 28, 2010

### kusiobache

Alright, I did it and I got 1.4 degrees.

My teacher said he remembers it being around 1 or 1.2 degrees so I think I'm correct. We are going to work it out during tomorrow class.

I'll explain my method later.