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Mathematics 23 Online
OpenStudy (kkutie7):

I need help. I'm not sure how to go about this problem.

OpenStudy (kkutie7):

More vector math. a. Two charges Q are fixed as shown (I'll draw a picture). A third charge -q (of mass m) is initially along the perpendicular bisector of the two Q's a distance x above the line connection them. It is then released from rest. Qualitatively describe the motion of the charge -q. Discuss force directions, accelerations, velocities, positions, etc.

OpenStudy (kkutie7):

|dw:1452988173360:dw|

OpenStudy (anonymous):

sorry am not in college but i can help tag the question to someone that knows way more than me @robtobey

OpenStudy (kkutie7):

I also know this \[|F_{Q,q}|=|\frac{-qQ}{4\pi \epsilon_{0} (\sqrt{x^{2}+D^{2}})^{2}}|\]

OpenStudy (kkutie7):

b. Determine the force as a function of x. Assuming that x<<D, determine the simple harmonic oscillation period of the charge -q.

OpenStudy (kkutie7):

|dw:1452988367355:dw|

OpenStudy (irishboy123):

you just need to resolve your force down/up the x-axis

OpenStudy (kkutie7):

Well I know the answers for both of a and b and i don't know how to get there. The force is similar to what I have calculated for one of the Qs on -Q. As for b i can't remember how to solve anything for harmonics i think i use this... \[T=2\pi\sqrt{\frac{m}{k}}\]

OpenStudy (irishboy123):

mmmm.... force between q and *either* Q is \(F = k \dfrac{qQ}{d^2} = k \dfrac{qQ}{x^2 + D^2}\) resolve that along the [vertical] x axis to get \(F = k \dfrac{qQ}{d^2} \cos \theta = k \dfrac{qQ}{x^2 + D^2}.\frac{x}{\sqrt{x^2 + D^2}}\) |dw:1452989094215:dw|

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