Does it matter if my x and y values are switched on 13-4 b) for Pc and Pd?

[Edited on April 16, 2007 at 10:53 PM. Reason : ]

[Edited on April 16, 2007 at 10:54 PM. Reason : rrr]

4/16/2007 10:43:15 PM

jbl4me, do you still need 14-2? I got it. I can't figure out 14-3 though.

4/16/2007 11:12:35 PM

So...I have them all now except 15-1 and 14-2 and 14-3

For 14-1 all you need is the formula (pexp(-(sigma/e)t)) correct?
What about 14-2?

For 14-3 Im guessing you just take the curl of G and then take the partial with respect to t?

I'm gonna get food...i'll be back

[Edited on April 16, 2007 at 11:15 PM. Reason : .]

4/16/2007 11:14:25 PM

For 13-4 b) does it matter if my x and y values for Pc and Pd are switched? I did it the same way in the worked example, but they are opposite in the sample test problem.

14-1
Just use the equation.

14-2
Use the equation from 14-1. You'll have 2 undefined variables (p and sigma), and you'll have two equations. Just solve one for p and the plug it into the other for sigma.

[Edited on April 16, 2007 at 11:20 PM. Reason : j]

4/16/2007 11:19:48 PM

Can anyone help me out with 15-4?

4/16/2007 11:57:50 PM

ninja, you stated:

"the direction of H is Uy b/c E is in teh Ux and propigates in teh Uy so Ux (cross) Uy = Uz"

Don't you mean the direction of H is uz?

4/17/2007 12:03:33 AM

^ 15-4 since H = E/n where n = sqrt(u/e) just evaluate for H.

14-2 I still cannot get... maybe cause my super annoying apartment mate is back and so i can't think

[Edited on April 17, 2007 at 12:08 AM. Reason : g]

4/17/2007 12:06:01 AM

14-2

p0(t) = 2 = p(t)*e^((-sigma / (20*8.854e-12))*t)
p0(t) = 2 = p(t)*e^(-5.647e9*sigma*t)..........eq 1

p1(t) = 1 = p(t)*e^(-5.647e9*sigma*(t+10^-6))...........eq 2

eq1......p(t) = 2*e^(5.647e9*sigma*t)
eq2......1 = [ 2*e^(5.647e9*sigma*t) ] * e^(-5.647e9*sigma*(t+10^-6))

Solve for sigma...I used my calc.

4/17/2007 12:10:32 AM

I'm only missing 15-1 and 14-3.

4/17/2007 12:17:33 AM

Yea me too...I THINK i could get 14-3, but I don't feel like futzing with it

4/17/2007 12:20:29 AM

on 15-1,

I know how they get their values...but I don't know why.
.0053 = 2 / 377 and .0133 = 5 / 377. I'm not sure why x and y are switched.

4/17/2007 12:22:38 AM

Ya rastaman8


H has a direction of Uz.

did anyone get 15-4 i have the right numbers but I don't know why ux has a negative in front of it.

4/17/2007 12:34:08 AM

I got 15-4:

I think the sample test problem answer is wrong...ux in H is supposed to be negative.

4/17/2007 12:35:55 AM

^ I agree its wrong in the notes. ux should be -

4/17/2007 12:37:51 AM

I'm really close on 14-3.

If you take the curl of G, you get:

90*e^(9y-6t) ux + 20*e^(4z-3t) uy

Integrate this to get F, and you get:

F = [-15*e^(9y-6t) + K1] ux + [-6.67*e^(4z-3t) + K2] uy

I'm not sure how to find both constants with only one boundary equation...

4/17/2007 12:39:14 AM

I'm glad you still have some energy left...lol

4/17/2007 12:42:34 AM

I have no clue how to get another equation.

I have one...K1 + K2 - 21.67 = 0.
which works since K1 = 15 and K2 = 6.67

But I don't know where to get the other one...

4/17/2007 12:45:21 AM

About time for bed...I'll be up early though and checking this...so feel free to try the rest of 14-3 or 15-1.

4/17/2007 12:58:42 AM

15-4 isnt wrong...

4/17/2007 9:11:50 AM

^ Then why isn't ux negative?

4/17/2007 9:43:26 AM

1/n *
| ux uy uz |
| 0 1 0 |
| -7.07 0 7.07|

1/n* ( ux * 1 * 7.07 + -uz * 1 * -7.07)

.0593ux + .0593uz

4/17/2007 10:04:58 AM

Where did the | 0 1 0 | come from?

And do you know how to do 15-1 or the end of 14-3?

4/17/2007 10:24:49 AM

i believe it was from saying p is in the pos y direction, its 1/n * uP x uE, i think...anyhoo off to test

4/17/2007 10:27:22 AM

that test blew major asshole

4/17/2007 12:34:59 PM

yea i failed it...

4/17/2007 1:03:25 PM

i had no clue about the moving charge particle, think it was 2 or 3

4/17/2007 1:16:00 PM

I couldn't do 2 or 3...

4/17/2007 1:29:10 PM

I just found E then F=qE

4/17/2007 1:52:47 PM

4/17/2007 3:46:58 PM