I figured it would be better to just start a thread for the sample questions.

I'm having trouble with 7-3 and 7-7 right now. Can anyone help me out?

3/19/2007 6:10:23 PM

Here is sorta what I did for 7-3:

First, find the Z looking in of the 2 that are in parallel. What you will find is that since the CharResistance = The Load Resistance, that the line is 'matched' to the load and such the the Z (R) looking in is 100ohms.

Second, those 2 are in parallel, so line one is looking at 50ohms.

Third, find Z1

100ohms*((50 + j100tan(2pi))/(100+j50tan(2pi))) Z1 = 50
Z1 is then just R1 = 50



Then by voltage division, 50 / 150 * Vs = 1.655

Ohms Law, 1.655/50 = 0.033A
Pavg = 1/2 i * v = .5*.033A*1.665V = 0.027W

Enjoy

<END 7-3>

I also have no Clue on 7-7
[Edited on March 19, 2007 at 7:45 PM. Reason : .]

[Edited on March 19, 2007 at 7:45 PM. Reason : .]

3/19/2007 7:44:14 PM

i cant figure out 7-7 or 10-4

:/

3/19/2007 7:49:10 PM

I think for 7-3, since all the lengths are even wavelengths, that they can be ignored, but i think that if one length was like .5 or something, it would through in a phase difference, making it look more like 7-4

3/19/2007 7:52:02 PM

7-4 is quite easy once you see what is going on. Convert the 2 to phazors, then Pavg = .5*v*i*

Anyone looked at 9-3 and 9-4? I can get 9-3 the distances correct, but I cannot get those voltages, they are way higher than what I can get.

9-4, why is the contour = to 20.25 and not just 20?



[Edited on March 19, 2007 at 7:59 PM. Reason : .]

3/19/2007 7:56:27 PM

I am having trouble figureing out any of the chapter 10 stuff. Any tips here?

3/19/2007 8:32:51 PM

I can get Vmin of 9-3, but I'm not sure about Vmax of 9-3.

Once you get the x of 9-3, just plug in the x in your original V(x) equation to get the voltage at that point. See page 15 of Notes Set 9 for an example.

When doing Vmax, will the x always be the low end of region R? If not, how do you calculate the x of Vmax?

3/19/2007 8:35:41 PM

In 9-4, the contour is 20.25 because

d = Q / (4pi e V)
d = 10*9 / 20 = 4.5

4.5 is the radius of the circle

(x-xo)^2 + (y-y0)^2 = r^2
r^2 = 20.25

[Edited on March 19, 2007 at 8:43 PM. Reason : ]

3/19/2007 8:40:40 PM

rastaman > I was messing up my voltages in 9-3 because I was not plugging in the 9e9 correctly. You should get the max voltage in 9-3 by:

V(x) = 2.76(9) / (x-0.5) + 1.14(9) / (10.5 - x)

Evaluate this function at xmax and xmin for the max and min voltages respectively.

Still having issues on 10-1 and 10-2

I tried Ex(x) = -1/3(x) but got nowhere. Any help here...?

[Edited on March 19, 2007 at 8:53 PM. Reason : info]

3/19/2007 8:50:16 PM

jbl4me,
I know how to get Vmin and Vmax, but how to do you get the x for Vmax? For Vmin, you set the derivative to 0 and solve for the roots. What do you do for Vmax? Thanks.

3/19/2007 8:54:09 PM

if you have a calculator, its easy enough just to graph it, other than that, both of your charges are outside of the range, if one were in it then your max would default to whichever charge was in range, since it would be infinite. so you just see which is higher, the far left or the far right, in this case it was the far left.

3/19/2007 9:14:55 PM

when you integrate dE(x) you get -1/3 x + C

and u know E is 0 when x = -2, so blug and chug

3/19/2007 9:18:23 PM

^ thanks got it...

Anyone looked at 10-4?

Looks like you would need to find an expression for a cylender then integrate it...

[Edited on March 19, 2007 at 10:02 PM. Reason : .]

3/19/2007 9:47:00 PM

What about 10-2? Since you don't have 2 regions, how do you satisfy both boundaries with only one ODE?

3/19/2007 9:54:54 PM

for 10-2 all I did was integrate the p(x)

so you wind up with Ex(x) = -10/2 x^2 + C.

then you take the fact that when x=2 E=0 and find that C = 4

that leaves you with the answer Ex(x) = 4 - x^2, -2..2

Now if I could just get 10-4...

3/19/2007 10:05:07 PM

I'm having a problem with 10-4 as well...here is what I am doing:

Vcyl = pi*(.004/2)^2*(.005)
p = Q / Vcyl

E = p / e...but I keep getting 3.183e-6

3/19/2007 10:11:14 PM

How do you get C = 4 if:

Ex(2) = 0 = -10/2 x^2 + C.

Wouldn't C = 20?

3/19/2007 10:19:36 PM

oops...i left out some stuff.

its -10e0x, but since permittivity is 5 the function you integrate would be -2e0x which comes out to be -x^2 + C

Sorry bout that tried to type with out looking back at notes.

Did anyone ever find out anything on 7-7? I had emailed Alexander Sunday night but still no reply as of now.

[Edited on March 19, 2007 at 10:27 PM. Reason : 7-7]

3/19/2007 10:23:05 PM

8-3...I got the homework correct by narrowing down 2 numbers until they worked...I won't have time on the test to do that though.

I know that Etotal = 9(10)^9 * [((5(10)^-9)/|r1|^2) * u1 + ((10(10)^-9)/|r2|^2) * u2] = 0
I know that y = -x + 10

When I put in r1, r2, u1, and u2, my calculator crashes because it can't handle that large of an equation. Am I missing a step? Where can I simplify this? Thanks.

3/19/2007 10:41:15 PM

What did you do for 8-4?

Use superposition...E = 17(9) / r^2 where r = 4.24...and so on for point 2...and repeat for point 1?

[Edited on March 19, 2007 at 10:49 PM. Reason : .]

3/19/2007 10:48:10 PM

I was waiting for 8-4 pending 8-3, haha.

3/19/2007 11:04:41 PM

How are you getting r when you don't know the final position yet?

3/19/2007 11:07:51 PM

I am getting r in 8-4...i have no clue in 8-3. It seems to me that in 8-3 the charge would have to be on the y=x line, but according to the answers it's not. I would think that if it was any other place, it would be repelled, since the charges are all 3 positive.

3/19/2007 11:10:34 PM

So on 8-4, I have E = EQ1 + EQ2...I'm not sure what to do with Q3...

3/19/2007 11:28:10 PM

What i did was add to each one. Since its superposition, E1 = (Q1 + Q3) and E2 = (Q2 + Q3)...of course they attract so you would have to get the signs right so that it sums properly. Then again, i didn't get the right answer...

3/19/2007 11:32:19 PM

If I turned E from Eq1 + Eq2 into a unit vector, I get the answer...so does the Q of charge 3 factor into this at all?

So, I did [Eq1 + Eq2] / [sqrt(Eq1^2 + Eq2^2)]...and got the answer.

[Edited on March 19, 2007 at 11:35 PM. Reason : ]

3/19/2007 11:34:04 PM

That must be the correct way, I reworked it here and got the same answers.

So all that we do not have solved are:

8-3, 7-7, 10-4

3/19/2007 11:44:15 PM

^ Same here.

3/19/2007 11:45:57 PM

I thought I had 9-6 correct, but I don't. Can someone give me starting point for 9-6?

3/20/2007 12:20:23 AM

For 9-6 you have a circle centered at (4, 3) Then just find the unit vector from (4,3) to (6,1). Thats what I did...anyways

3/20/2007 12:27:17 AM

^ jbl4me, you've been a great help...I'm going to work on 7-7, 8-3, and 10-4 in the morning. If I figure any of them out, I'll post them here before class time tomorrow.

3/20/2007 12:38:49 AM

10-4:

You need charge density which is charge per unit volume.

Volume for cylinder: (pi * r^2) * length

Charge density(p) is therefore 10^-12 / volume

div(E) = p / epsilon

Epsilon = 5*epsilon0, with epsilon0 being 8e-12


I got the right answers for 7-7 and 8-3 but I think my logic is a little shakey.

[Edited on March 20, 2007 at 3:40 AM. Reason : ]

[Edited on March 20, 2007 at 3:40 AM. Reason : ]

3/20/2007 3:39:20 AM

I got 8-3.

Do the same thing as in 8-4. Just do (EQ1 + EQ2)=0. r = sqrt[(x-0)^2 + (y-10)^2] for u1, etc...

Get to where you have an equation in the form [...ux + ...uy] + [...ux + ...uy] = 0

Now, forget about the uy's...and add together the ux's.

Replace y with -x + 10 (this is the line that Q1 and Q2 are on).

Solve for x.

Now, y = -x + 10.

3/20/2007 7:46:00 AM

^^neolithic, what did you do for 7-7?

3/20/2007 7:50:40 AM

Alright:

R = L / (A * sigma) where L is length and A is area of the wire

When you do the calculations you get the each wire has a resistance of 43.9 ohms, that combine to give you a total series resistance of 87.8 ohms. I think modeled the circuit as the characteristic line resistance in series with the load. All you do then is simple voltage division and Pave = V^2 / R


Could you tell me exactly how you did 7-4?

3/20/2007 8:16:54 AM

7-4:

v phasor = 10e^(...) + 6e^(...)
i phasor = 10/100e^(...) + 6/100e^(...)

vphasor x iphasor*

Pavg = (1/2)*Re(vphasor x iphasor*)

Page 11 in Notes Set 7 does the exact same problem with different numbers.

3/20/2007 9:00:07 AM

Yes, but when I expand I get a cos instead of a jsin term. So I have 1.36 + cos(2*pi / 30 x). Am I just multiplying out incorrectly?

3/20/2007 9:05:40 AM

It doesn't matter what you get...you are only taking the real part in the end.

Edit: I just saw you didn't get a j at all...you are just multiplying incorrectly.

[Edited on March 20, 2007 at 9:16 AM. Reason : ]

3/20/2007 9:15:35 AM

When I multiply out, I get

0.64 + j1.2sin(2Bx)

3/20/2007 9:16:55 AM

Well...it's all over but the cryin'...how'd every body do?

I think i got them all but the last one.

3/20/2007 2:17:35 PM

Barring stupid errors, I think I did alright. I was able to able to do all of the problems.

3/20/2007 3:33:43 PM

"Barring stupid errors, I think I did alright. I was able to able to do all of the problems."

3/20/2007 6:07:49 PM

Quote :

"Barring stupid errors, I think I did alright. I was able to able to do all of the problems."

3/20/2007 10:17:25 PM