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By dlgn
#188429
Okay, so I have a question about electrophysics.

On my second physics midterm, there was a question about an electron and proton at infinity at rest that accelerate toward each other and eventually meet.

The first part of the question was about which has a greater acceleration. Well, the electron does, because they exert an equal force on each other but the electron has a smaller mass.

The second was about which travels further before they meet. The electron again, because they have the same initial velocity (0 m/s) but the electron has a greater acceleration.

The third and last part of the problem is what confused me: which has a greater kinetic energy at the moment they meet?

Okay, so first try a work-based analysis (which is what the exam recommended, and the one I ended up going with in the end). By conservation of energy, the change in kinetic energy is equal to the work, which is equal to force*displacement. Well, they have the same force but the electron has a greater displacement, so it would have more final KE, right?

But then I did a potential energy analysis, and found that since they undergo the same change in potential energy (with an initial PE of 0 and a final of -ke^2/r_final), they must undergo the same change in kinetic energy, meaning they have the same KE at the end. What? I mean, this makes more sense from a conservation of energy standpoint (where did the extra work done on the electron come from, anyway?) but it's inconsistent with the work analysis. So, what's going on here? Cube? Anybody?

Thanks,
~dlgn
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By Aelcalan
#188430
Mass.
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By dlgn
#188438
Okay, I think I figured it out: since the force is non-constant, I can't use the F*x formula, and instead have to integrate. I did it in a bunch of different ways (including one that lead to an interesting problem about the time for the electron to move a certain distance that I haven't solved yet), and got the same answer.

Image

~dlgn
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By Daegon_Phyn
#188439
Look at your force integral. It's simply the integral of Fdx from infinity to 0. Well not 0 because electrons and protons are physical objects and have a radius. So let's say to the radius of the electron (that's not quite right either because a proton is made up of quarks and quantum effects come into play). The force goes like one over distance squared. So integral of 1/x^2 from infinity to some radius, r_e, is -1/r_e. And then the constant term is just k q_e q_p. Overall term is positive and independent of mass and distance traveled.

This property, that the distance traveled doesn't affect the work done, is what makes electromagnetism a "conservative force". Gravity is also a conservative force. Basically, it means that the work done is independent of the path traveled.
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By dlgn
#188449
Daegon_Phyn wrote:Look at your force integral. It's simply the integral of Fdx from infinity to 0. Well not 0 because electrons and protons are physical objects and have a radius. So let's say to the radius of the electron (that's not quite right either because a proton is made up of quarks and quantum effects come into play). The force goes like one over distance squared. So integral of 1/x^2 from infinity to some radius, r_e, is -1/r_e. And then the constant term is just k q_e q_p. Overall term is positive and independent of mass and distance traveled.

This property, that the distance traveled doesn't affect the work done, is what makes electromagnetism a "conservative force". Gravity is also a conservative force. Basically, it means that the work done is independent of the path traveled.
I know, I just wanted to be absolutely sure.
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By SkullRose
#188457
It's been a while so here's some random pictures of stoof :3
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We-have-puddings.jpg
We-have-puddings.jpg (80.98 KiB) Viewed 11225 times
Poor-apatosaurus.jpg
Poor-apatosaurus.jpg (51.62 KiB) Viewed 11225 times
IMG_1114175444476931.jpeg
IMG_1114175444476931.jpeg (24.08 KiB) Viewed 11225 times
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By dlgn
#188459
Does anyone have any idea how to evaluate this integral? Because I and my sciency friends can't figure it out, and neither can WolframAlpha.
Attachments
WolframAlpha--integral_from_0_to_t_of_sqrtax-a22bx_dx--2014-05-16_1334.png
WolframAlpha--integral_from_0_to_t_of_sqrtax-a22bx_dx--2014-05-16_1334.png (3.33 KiB) Viewed 11207 times
By Skylord_Fox
#188460
dlgn wrote:Does anyone have any idea how to evaluate this integral? Because I and my sciency friends can't figure it out, and neither can WolframAlpha.
Divide by zero.
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By Daegon_Phyn
#188463
dlgn wrote:Does anyone have any idea how to evaluate this integral? Because I and my sciency friends can't figure it out, and neither can WolframAlpha.
Just going to do it as an indefinite integral. First simplify so your integral is sqrt(a/(2b))*sqrt(x-a)/sqrt(x)*dx. Then do integration by parts where u=sqrt(x-a) and dv=dx/sqrt(x). You get an integral that is sqrt(x)/sqrt(x-a)*dx, but you can find that in an integral table. The final result is sqrt(a/(2b))*[sqrt(x(x-a))-a*ln(sqrt(x)+sqrt(x-a))]. You can see if you plug in x=0 the last term is ln(sqrt(-a)) which means a has to be negative and that also means b has to be negative so the leading term isn't imaginary.
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