Question

BETTER FORMATTED QUESTION: Hello! ^^ Here is the reference image: http://imgur.com/M6du4j2 And here is the question! Two infinitely long conducting wires are located at x= - R and x=R, as shown below. The wires both have linear charge density of + λ , and are parallel to the y - axis. A small charged particle with mass m and electric charge of +q, is released from rest at a small distance of a (a< 12 years ago

Answer 1

c)What is the maximum speed of this particle? d)This particle will oscillate around the center at x=0. How long will it take the particle to come back to point A? (5pt) (Hint:You need to use what you have learned about simple harmonic motion in PHY2048 and use the approximation of€R2−a2≈R2) (This hint is only for part d, and should not be used for part a,b,c)

Answer 2

(R^(2)-A^(2)≈R^(2))*fixed typo, and a big thanks to wolfe7 for his help in me posting this correctly! I look forward to hearing from you guys!

Answer 3

What sorts of attempts have you made on the problem? Are you still on part A?

Answer 4

I believe I can find part A using the relationship F=kq1q2/r^2 and applying that to A v=KQ/R sort of formula, but I just don't know how to start this, yes, I'm stuck on part A

Answer 5

Perhaps find x ad y components and then summing to find magnitude? I just dont know where the numbers come in or how they do

Answer 6

Are you in a vector-calculus E&M class or is this more of an introductory course to electromagnetism?

Answer 7

physics w/calc2, with first test coming up, stressful question to test how prepared we are

Answer 8

So yes, introduction to electromagnetism

Answer 9

OK. You'll need to use the integral form of the electric field here since it's a continuous charge distribution and not a discrete one:
\[\vec{E}=\frac{1}{4\pi \epsilon_0} \int\limits\limits \frac{1}{r^2}dq\ \hat{r}\]You've seen this form before, yes?

Answer 10

Yes, that appears to be Gauss's law with 1/4pieo as constant k=9x10^9 with (1/a^2 - 1/b^2) where a and b are the end of the integral and r hat as the direction

Answer 11

where the difference of the fractions are the integration of the integral and dq as some infentecimally small area of a Gaussian surface

Answer 12

Buuut I'm not exactly sure how this ties into to solving for a, I am confused

Answer 13

So solving for the electric field will also solve for the net force (magnitude and direction) on point A? because they are on in the same, but I've no idea what to plug-in for the numbers...so then would the expression for the electricfield you posted be enough for just part a?

Answer 14

one in the same*

Answer 15

im rly lost

Answer 16

Would this help? http://www.public.iastate.edu/~lhodges/112/Topic-4.pdf

Answer 17

not really >.> i need drect application not mountains of paper to read through

Answer 18

No no read the first 1 and half page. Because even I have to re-study this. I kinda of rushed through the semester :/

Answer 19

and after skimming the majority of it, doesn't look worth reading through to answer the question, ill probabbly come out of it more confused

Answer 20

hmm lemme try

Answer 21

Well just that the distance and force proportion or something.

Answer 22

im not getting anything from this T.T it pertains to thi yes, but doesnt help me too much

Answer 23

Ah, yeah, with a simplification, this can be solved using Gauss' law with a cylindrical Gaussian surface.

You can solve for the electric field contributed by each wire at point A using Gauss' law. To find the force, just:\[\vec{F}=\vec{E}Q\]

Answer 24

Are you comfortable with that much, or do you need a nudge in the right direction?

Answer 25

Ehhhh yah I need more than that

Answer 26

OK, you've talked about Gauss' law and how you can use symmetry to simplify problems like this, using:\[\int\limits \vec{E}\cdot d\vec{a}=\frac{q_{enc}}{\epsilon_0}\]So first you need to define your Gaussian surface. In this case, because of the shape of the wire, you can use a Gaussian surface which is a cylinder.

Answer 27

yes, the formula you posted looks very familiar to me, and selecting a gausian surface makes sense. a cylindrical surface that is not infinitely long like the wires, but a surface that surrounds an area from say a to b, but i think a picture would rly help me see things clearly, not sure over which region the surface should cover

Answer 28

perhaps a surface like that is what i mean