Question about symmetry with potential fields.

[mateomy]

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?

 

[gneill]

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?

 

[mateomy]

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.

 

[gneill]

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

 

[mateomy]

I was attempting to try it with conservation of energy.

$$\frac{1}{2} m(v^2)_i + q\phi_i = \frac{1}{2} m(v^2)_f + q\phi_f$$

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?

 

[gneill]

If I may suggest, why not calculate the work done in moving a charge q through a potential difference V?

 

[mateomy]

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.

 

[gneill]

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.

 

[mateomy]

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.

 

[gneill]

The electric field between (sufficiently large) plates is essentially uniform, so yes, halfway between the plates the potential should be half as well.

 

[mateomy]

Great, thanks.