Question about symmetry with potential fields.

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The discussion revolves around determining the final velocity of an electron traveling between a cathode and an anode with a potential difference of 300V. It is suggested that by symmetry, at half the distance to the anode, the potential could be assumed to be 150V. The conversation highlights the importance of the potential difference in calculating the final velocity using conservation of energy principles. It is concluded that since the electric field between large plates is uniform, the potential at the midpoint can indeed be considered half of the maximum potential. This assumption simplifies the problem and allows for easier calculations of the final velocity.
mateomy
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Quick question (I think anyway)

I'm currently trying to solve a problem that essentially is asking for me to find the final velocity of an electron that is traveling between a cathode and an anode of potential difference 300. At half the distance to the anode can I assume by symmetry that it would be half the total potential difference i.e. 150V?
 
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mateomy said:
Quick question (I think anyway)

I'm currently trying to solve a problem that essentially is asking for me to find the final velocity of an electron that is traveling between a cathode and an anode of potential difference 300. At half the distance to the anode can I assume by symmetry that it would be half the total potential difference i.e. 150V?

You could... How do think that will help you?
 
Well if I know the potential at one end is say, 300V and I arbitrarily set the cathode side to 0 for sake of convenience -knowing that it is only the difference in potential that is important- and I want two values; one half way and one full displacement, symmetry can help a lot. If I make that argument I can conclude that half the distance would be half the potential.
 
mateomy said:
Well if I know the potential at one end is say, 300V and I arbitrarily set the cathode side to 0 for sake of convenience -knowing that it is only the difference in potential that is important- and I want two values; one half way and one full displacement, symmetry can help a lot. If I make that argument I can conclude that half the distance would be half the potential.

Oookaaay... still not seeing the utility of that for solving a problem which is otherwise quite straightforward. However, I am prepared to be amazed :smile:
 
I was attempting to try it with conservation of energy.

<br /> \frac{1}{2} m(v^2)_i + q\phi_i = \frac{1}{2} m(v^2)_f + q\phi_f<br />

Where v_i is 0, and \phi_i is 300 and q is the charge of the electron. Manipulation of this equation, isolating final velocity I get a difference of potentials under a square root. That's why I have to make an assumption based on symmetry. Does that work?
 
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If I may suggest, why not calculate the work done in moving a charge q through a potential difference V?
 
Hmmm, seemed to have overlooked such an obvious way of doing it. I will get right on that. In any case does my argument have any validity?

Thanks for the pointers by the way.
 
mateomy said:
Hmmm, seemed to have overlooked such an obvious way of doing it. I will get right on that. In any case does my argument have any validity?

I'm afraid I just can't picture a way forward with it; to me it just seems to divide the problem into two problems with the same unknown quantities.
 
I don't know. It only leaves one unknown being \phi_f with the obvious exception of the final velocity. Which brings me back to the initial question, can I assume the potential would be half its max (or min) at half the distance? If it is, then I only have the one unknown of the final velocity. Maybe I'm overlooking something, but it seems pretty easy to solve that way as well so long as I can make that assumption with respect to the potential.
 
  • #10
The electric field between (sufficiently large) plates is essentially uniform, so yes, halfway between the plates the potential should be half as well.
 
  • #11
Great, thanks.
 
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