# Thread: Velocity

1. ## Velocity

Does anyone have an equation for calculating the velocity of an electron from the anode of a crt to the cathode? Any help would be greatly appreciated.

2. Originally Posted by Dumkoff
Does anyone have an equation for calculating the velocity of an electron from the anode of a crt to the cathode? Any help would be greatly appreciated.
(voltage difference between anode and cathode)(charge of the electron) = potential energy of the electron

as the electron accelerates from anode to cathode, the potential energy changes to kinetic energy.

3. skeeter, wouldn't the distance between anode and cathode be a factor in the electron's final kinetic energy?

4. Originally Posted by Dumkoff
skeeter, wouldn't the distance between anode and cathode be a factor in the electron's final kinetic energy?
yes ... assuming a uniform electric field E, V = Ed, where d is the distance.

5. Ok I've been unable to do it. If anyone knows, what would the equations be to calculate the impact velocity of an electron if the distance from anode to cathode is one meter and the potential difference is 35Kv

6. Originally Posted by Dumkoff
Ok I've been unable to do it. If anyone knows, what would the equations be to calculate the impact velocity of an electron if the distance from anode to cathode is one meter and the potential difference is 35Kv
Since you're given the potential difference the distance is irrelevant.

The gain in K.E. is $\displaystyle (35 \times 10^3) \cdot (1.602 \times 10^{-19})$ Joules.

And K.E. $\displaystyle = \frac{1}{2} m v^2 = \frac{1}{2} \cdot (9.1095 \times 10^{-31}) \cdot v^2$.

Equate these two and solve for $\displaystyle v$.

7. So the terminal velocity for 0.5 meters is the same as 1 meter? One would think that accelerating over a longer time period would produce a higher terminal velocity

8. Originally Posted by Dumkoff
So the terminal velocity for 0.5 meters is the same as 1 meter? One would think that accelerating over a longer time period would produce a higher terminal velocity
If it only accelerated over 0.5m then the potential difference would only be 0.5/1 = 1/2 of 35kV (assuming it was in the same electric field). You need to understand the meaning and relationship between electric field and potential difference.

9. Maybe Im not looking at the problem in the correct way. In a TV crt the cathode emits electrons that raster across the screen. I'm trying to calculate the difference in velocity of an electron that hits on the left edge of the screen as compared with one that hits in the center.