Voltage Used to Fire an Electron

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burnst14
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Homework Statement


If an electron travels 0.210 m from an electron gun to a TV screen in 29.7 ns, what voltage was used to accelerate it? (Note that the voltage you obtain here is lower than actually used in TVs to avoid the necessity of relativistic corrections.)


Homework Equations


Kinematics equations, but past that, I don't have one.


The Attempt at a Solution


I have calculated the acceleration and change in PE, as prompted by WebAssign, but I only have an equation for ΔV. I tried assuming the voltage at the TV screen was zero, but the answer was incorrect.

a = 4.76E14 m/s2
vfinal = 14,141,414,14 m/s
ΔPE = -ΔKE = -6.44E-24 J
ΔV = 4.03E-5 V
 
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I agree with the acceleration you found. Velocity, I am less sure. I get 14141414.14 m/s, and if the final comma in your figure was actually a period, then we have the same value. But I obtain a completely different number for the change in energy.
 
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The electron is accelerated in the electron gun and not between the gun and the screen.
You cannot (and don't need to) calculate the acceleration in the gun.

Just find the voltage necessary for the electron to leave the gun with a speed so that it can travel the given distance in the given time.
 
Ah, yes indeed. Somehow I imagined that the TV screen was the anode, which is of course not true.
 
Alright, I recalculated change in energy and obtained 9.11E-17 J. Yes that comma is supposed to be a period. All right, now. I have no idea what equation I would use to find that voltage. I looked through my notes and didn't see one even close, except that ΔV one I tried to use. Also, I don't know this word, anode.
 
Okay, I got the answer right with my new potential energy. I must have typed something into my calculator wrong. I'm still kind of confused about why that's correct though. Is that "Note" at the end of the problem statement basically saying that there is no voltage at the TV to slow down or speed up the electron en route?
 
The anode is the positive electrode. Since you got the correct answer, apparently you do have to assume that the TV screen is the anode. Which is not true on real TVs, because the accelerating voltage is usually quite high, which, even if is not lethal, would be quite unpleasant to unsuspecting customers.
 
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Alright thanks. Plenty more homework to do, so I'll probably be back.