[quote]asusvenus wrote:
mmf?
oh wait…
[/quote]
Particle 2 has a slutty little ass
You made 2 mistakes that I can see: you tried to solve without breaking down into component, and you complicated the matter by thinking about it in terms of forces instead of trying to sum the E fields to 0.
Using the attached diagram E(x) = E1(x) + E2(x) = kq1/(d1^3) + kq2/(d2^3) = 0, then substitute in known values and solve for q3(x). Repeat the same procedure in the y direction and you should obtain your coordinates.
[quote]JLu wrote:
You made 2 mistakes that I can see: you tried to solve without breaking down into component, and you complicated the matter by thinking about it in terms of forces instead of trying to sum the E fields to 0.
Using the attached diagram E(x) = E1(x) + E2(x) = kq1/(d1^3) + kq2/(d2^3) = 0, then substitute in known values and solve for q3(x). Repeat the same procedure in the y direction and you should obtain your coordinates.
[/quote]
No, no, no. That’s all wrong. Don’t listen to JLu. Use this diagram instead.
[quote]super saiyan wrote:
[quote]JLu wrote:
You made 2 mistakes that I can see: you tried to solve without breaking down into component, and you complicated the matter by thinking about it in terms of forces instead of trying to sum the E fields to 0.
Using the attached diagram E(x) = E1(x) + E2(x) = kq1/(d1^3) + kq2/(d2^3) = 0, then substitute in known values and solve for q3(x). Repeat the same procedure in the y direction and you should obtain your coordinates.
[/quote]
No, no, no. That’s all wrong. Don’t listen to JLu. Use this diagram instead.
[/quote]
The eyes aren’t beady enough, no Canadian has eyes that fucking large.
[quote]DBCooper wrote:
[quote]Sick Rick wrote:
Why is everyone bitching? People post math questions on here all the time and they don’t get flame-raped.[/quote]
Go to the SAMA forum and read the “First Date at 19” thread. I think that’s what it’s called, something along those lines. Maybe that was Herc’s. Anyways, just look for his post in SAMA. He started it a few weeks ago. It will explain everything.[/quote]
Haaaaaaha. Ok I get it.
Guy’s obviously a troll; mormons don’t believe in physics.
[quote]jdinatale wrote:
I’m not asking you to do my homework, only to help me find where I’m messing up.
Here is how one guy did it, but I couldn’t follow his notation:
http://answers.yahoo.com/question/index?qid=20080219182823AAcGWwh
This is how I attempted it:
If particle 3 is to remain stationary, then the force on particle 3 from particle 1 must equal the force from particle 2 on particle 1.
Using coulombs law to state this:
F=[K(q1)(q3)]/(r^2) = [K(q2)(q3)]/(r^2)
But, particle 3 has to be to the right of particle 1, if it was in the middle between particle 1 and particle 2, then it would be pulled to the left towards particle 2, because particle 1 would repel particle 3 to the right, and particle 2 would pull particle 3 left.
So to make up for this in our equation we make the distance from charge 2 to charge 3, “R+L” where r is the distance between charge 1 and charge 2:
F=[K(q1)(q3)]/((r)^2) = [K(q2)(q3)]/((r+L)^2)
Now we can place some numbers in this equation, and solve for L:
F=[(8.99x10^9)(3.5x10^-6)(5.5x10^-6)]/((.0024)^2) = [(8.99x10^9)(-4.0x10^-6)(5.5x10^-6)]/((.0024+L)^2)
Coulombs constant and Q3 cancel from the equation, so I’m left with
F=[(3.5x10^-6)]/((.0024)^2) = [(-4.0x10^-6)]/((SQRT(.0026)+L)^2)
(I know that the two original particles are SQRT(.0026) m apart)
So now we have an equation with only one unknown, L, so I solved the equation for L and got that L was .03765662 meters.
So we now know that particle 3 is 3.765662 CM from particle 1.
We can add that to the original coordinates of particle 1, to find out where particle 3 is:
the angle that these three particles lie on is -11.3
x= 3.765662cos(-11.3) + 3 cm = 6.692663 CM
y= 3.765662sin(-11.3) + .5 cm = -.2378669183 CM
However the online grader says that those are both wrong. AND I ONLY HAVE ONE ATTEMPT LEFT TO GET CREDIT[/quote]
Tell your sister I said,Hi.
[quote]JLu wrote:
You made 2 mistakes that I can see: you tried to solve without breaking down into component, and you complicated the matter by thinking about it in terms of forces instead of trying to sum the E fields to 0.
Using the attached diagram E(x) = E1(x) + E2(x) = kq1/(d1^3) + kq2/(d2^3) = 0, then substitute in known values and solve for q3(x). Repeat the same procedure in the y direction and you should obtain your coordinates.
[/quote]
It is extremely important that you learn to solve problem by looking at the E fields, it makes life much simpler. Later you will have do the same using potential.
Bellmar
[quote]Bellmar wrote:
[quote]JLu wrote:
You made 2 mistakes that I can see: you tried to solve without breaking down into component, and you complicated the matter by thinking about it in terms of forces instead of trying to sum the E fields to 0.
Using the attached diagram E(x) = E1(x) + E2(x) = kq1/(d1^3) + kq2/(d2^3) = 0, then substitute in known values and solve for q3(x). Repeat the same procedure in the y direction and you should obtain your coordinates.
[/quote]
It is extremely important that you learn to solve problem by looking at the E fields, it makes life much simpler. Later you will have do the same using potential.
Bellmar[/quote]
Now I have taken shrooms and sat in a field before but I never would have thought to take E, sit in a field and do my math homework. Interesting idea though. Does it work with other types of homework as well?
V
I personally would argue that the uncertainty principle should prevent anyone from knowing both the location and velocity of said particle thus rendering the problem physically impossible.
“coordinates of two charged particles held fixed in an xy plane” ← these are impossible initial conditions.
[quote]Nards wrote:
I’ve been a Star Trek fan for close to 24 years…I became one when I was 13. There was a noticeable decline in my interest that started the very moment I lost my virginity. I still like it, but certainly not as strongly.
That day was yesterday.[/quote]
LMFAO! must be the Shirtless Kirk cologne - set phasers to stunning.
[quote]PimpBot5000 wrote:
Also, Coulomb is a dick.[/quote]
Agreed. Always poking his nose in other people’s illegal activities.
I am disappointed with you JD
I tought you were the next Nicolas Tesla and now you need help with this introduction stuff