1. Physics E&M

I don't know what I'm doing wrong.

A proton is projected in the positive x direction into a region of uniform electric field E = -5.30e5 i N/C at t = 0. The proton travels 6.90 cm as it comes to rest.

Acceleration = -5.07e13

Initial Velocity = 2644779

Determine the time interval over which the proton comes to rest.
?????

I don't understand.

xf = xi + vi*t + 1/2*a*t^2

0.069 = 0 + 2644779*t - 2.535e13*t^2

-2.53e13*t^2 + 2644779*t - 0.069 = 0

a = -2.535e13
b = 2644779
c = -0.069

x = [ -2644779 +/- sqrt(2644779^2-4*-2.535e13*-0.069) ] / (2*-2.535e13)

gives imaginary solution... using numbers that are correct/given.

This is just a simple quadratic problem, right?
Or am I missing something and I'm somehow supposed to use the E field? Professor cancelled the second lecture (which is where we were supposed to discuss E fields and whatnot), but this homework is still due tonight.

bonus:
how would one go about solving a problem like this? The textbook is confusing me and I don't quite understand where some variables come from.

A +2 microCoulomb charge is located at the origin. What is the magnitude of the E field from theis charge at x = 2 m (on the x-axis)?

thanks @xXiris;!

EDIT:
I kept getting 2.61e-8 when I messed around with the signs, and I got frustrated so I started guessing. Around 4.8e-8 webassign said I was within 10% of the answer.. so I looked back and wondered if I doubled 2.61e-8. It worked.

The answer is 5.22e-8. I realize I was off by a multiple of two and I don't have the brainpower left tonight to see where I went wrong. If anyone else wants to point out the mistake while I sleep, be my guest. I'll look it over tomorrow.

2. Re: Physics E&M

Two things:

1) you're not off by a factor of 2 (which implies a conceptual misunderstanding); the impossible answer (imaginary solution) is supposed to be the correct one, you just made arithmetic errors (namely too much rounding).

2) Well actually I said both things already. The headache is due to your rounding the acceleration and the initial velocity to those values.

Assuming that the acceleration and initial velocity are not given by the problem (because they're what's causing the headache), here's the solution.

Spoiler

The quadratic is supposed to have one solution only, and the positive discriminant comes out of rounding errors (b^2 - 4ac = 6.29 while either of b^2 or 4ac is of 12 orders of magnitude away). When you rounded the acceleration and initial velocity to what you have in the OP, you made the discriminant negative.

@Marksman Bryan

Bonus: The field is just a force without a victim, so calculate the electric force on another charge (make up one), if you have to, then divide by the magnitude of the victim charge. Otherwise, E = kq/r^2.

3. Re: Physics E&M

I re-read the textbook and Matt explained the field problem to me, so I'm pretty sure I understand that now.

The possibility of it being a rounding error crossed my mind for a brief second while I was lying in bed last night, but dismissed it because I figured I wouldn't have been told the acceleration/initial velocity were right.

Thanks a ton @Kalovale;! I know from now on that a problem that requires this much accuracy needs more precision when it uses calculated variables.

4. Re: Physics E&M

I know you've already got an answer for both things, but I'm going to stick my neck in anyway, but you didn't really need to do the whole quadratic thing.

If you know the initial velocity (2644779m/s), the final velocity (0m/s) and distance travelled(0.069m) for a constant acceleration (-2.535e13m/s^2), you could just use d=(u+v)*t/2 and come up with a much simpler equation where you less likely to make a mistake calculating it (This happens an awful lot, it's best to avoid the problem wherever possible )

t =2*d/(u+v) = 2*0.069/(2644779)=5.22e-8 s

And the second bit is fairly straightforward too.

E field due to a charge Q is just Q/(4*Pi*permittivity of free space*r^2), and in this case, r =2m. Occasionally, they might make it harder, and put it at some point (x,y,z) from the origin, in which case you just use Pythagoras theorem to find the radius and away you go. Multiple charges add like vectors.

5. Re: Physics E&M

Yeah, don't know why I couldn't do the E field problem. After Matt told me kq/r^2, I looked in the textbook again and it was pretty clear that I should have been using that.

I've had one physics class this semester; guess I'm just not back in the groove yet. :(

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