capa 5 - Phys 1120 CAPA#5 solutions 1 This problem gives...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Phys 1120, CAPA #5 solutions. 1) This problem gives you a configuration (parallel plates, a given distance d apart) which is quite familiar by now. We accelerate a particle (with given mass and charge) across a given voltage V. I ask for kinetic energy. There are many ways you might think to proceed, but it seems to me that the simples, when asked for energy, is to think about energy! Here we know that the particle is changing voltage, which means we immediately know that its potential energy changes, since Delta U = q Delta V. That's the BASIC defining formula for voltage, all others come from it! If you change voltage by 200 Volts, and you have a charge of, say, 2e, then you change potential energy by q*Delta V = 2e*200 V. You can plug in e in metric units, or you can do the nice trick and see that this is just 400 eV without needing any calculator. Now, that's the DROP in potential energy of the particle. But energy is conserved, so if it loses this much potential energy, it must GAIN exactly this much kinetic energy, right? So we're done, it started with zero kinetic, so the gain is going to also be the final kinetic energy. You could just type "400 eV" (or whatever your number for voltage was, times 2 since the charge of this particle was 2e) and you're done. There was more information here than you needed (mass, and distance), which will happen more and more often as we move on - a big part of solving real problems (as you move beyond "coursework") is figuring out what information you NEED and what is superfluous! Note that you could solve this problem thinking about force, and acceleration, and old 1110 kinematics equations, I leave it to you to think about that if you like. But the energy approach is sure quick and easy when it works! 2) I hope we all have the same figures (!?). My fig 1 had +q and -q on the top, -q and +q on the bottom. (My fig 2 had -q and -q on the top, +q and +q on the bottom. ) Think about voltage first, it's easier. At point a, we're equidistance from +q and -q on the top, the voltage arising from the two of them therefore cancel. (kq/r + k(-q)/r). Same happens for the bottom pair. Adding all four, we have V=0. At point b... it's the same story! We have V = kq/r1+k(-q)/r1 (where r1 is the distance from b up to those upper charges) + k(-q)/r2 + k(q)/r2 (where r2 is the distance from b to the nearer lower charges). Do the sum - these still cancel to give 0. So V(b)=0. At c, we're equidistant from -q and +q on the left, they cancel. We're equidistant from -q and +q on the right, they cancel, and the sum of all four gives V(c)=0. Finally, at d, we are CLOSER to the two +q's, and FARTHER from the two -q's, so at last V will not cancel out. (The nearby +'s win, in terms of overall sign, because they're closer: V(d)>0.) That answers half the questions. Now for E fields: You have to think about the "arrows" from ALL FOUR corners, at the points in question. Just do them one at a time!...
View Full Document

Page1 / 6

capa 5 - Phys 1120 CAPA#5 solutions 1 This problem gives...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online