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E&m problems

  1. Feb 19, 2006 #1
    i have a few problems that i was struggling with.

    Projectile motion:
    An electron is fired in an electric feild E=-720j. Its launched with a velocity of 9550 m/s. if you want the x distance it travels to be 1.27mm, what are the two values of theta that will work? how long does it take?

    so, like i do all projectile problems, i found t using the x components, (which i left in terms of theta,) because there's no force in x. then i plugged t into y_f= 1/2at^2 +v_ot. my problem was that i was left with cos and sin's and i didn't know what to do with them. i couldn't easily solve for theta. and at any rate, i wasn't really sure how to use the electric feild. when i thought about gravity fields, i thought that i should be using energy or something, and i just got confused.

    SHM of charge ring:
    i forget the wording of this one, but its just like, show that the frequency of a particle moving on the axis of a ring is [tex]f=\frac{1}{2\pi}*(\frac{kQq}{ma^3})^1/2[/tex]

    i tried starting with the electric feild, and getting the force from that. when my prof did an example with the spring force in class, he did this really weird thing i can't follow now. he was like f=1/s and so if F=kx=ma then k/m*x=d^2x/dt^2, and what function looks like its second derivative? sin. and then we have the constant k/m, which must be frequency squares since sin(ax)'s second deriv is a^2sin(ax). and thats how he found the frequency for the spring. i have NO idea how to do that with the electric force. i tried working backwards too, but i couldn't really make sense of it because there's x's in the denominator of the force and not the frequency, and yeah, i'm stuck.

    The last problem i'm stuck on is this-
    Electric quadrapole:
    whats the electric feild along the y axis if there's a quadrapole on the x axis, with charge q at (-a,0), -2q at (0,0) and q at (a,0)?

    so i started with writing the feild generated by each. knowing that the x components of the two q charges would cancel, and their y's would be equal. so i got two terms, and i simplified and used the binomial theorem, and i couldn't get it right. but i think i'm going to try this one again.

  2. jcsd
  3. Feb 20, 2006 #2
    i really do need help on the harmonic motion one. i'm super stuck. i'm doing the first one with work-energy theorem, and i think i'm just screwing up the math in the third one, but the second one i'm really clueless on.
  4. Feb 20, 2006 #3


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    Rewrite everything in terms of [tex] tan (\theta) [/tex] (you will need to use [tex] {1 \over cos^2 \theta} = 1 + tan^2 (\theta) [/tex]). Then you wilh have a quadratic formula for [tex] tan (\theta) [/tex] . Solve and then take the inverse tan of the two roots.

    That's right. All you have to do is to show that the electric force on the point charge has the form [tex] - C~ y [/tex] where C is a constant. Then, y(t) has to follow simple harmonic motion with [tex] \omega [/tex] given by [tex] \omega = {\sqrt {C \over mass}} [/tex]. So all you have to know is the electric field of a ring along its axis and that times the charge of the point charge will give you the force.
  5. Feb 20, 2006 #4
    For your first question:
    [tex]\vec{a} = \vec{F}/m_e = \frac{-e(-720\vec{j})}{m_e}[/tex]

    [tex]at = v_0 \sin(\theta)[/tex]
    [tex]\Delta x = v_0 \cos(\theta) t[/tex]
    substituting t with other stuff
    [tex]\Delta x = v_0 \cos(\theta) v_0 \sin(\theta) / a[/tex]

    I think what you needed was trigonometic identity [tex]2 \sin(\theta) \cos(\theta) = \sin(2 \theta)[/tex]

    Use it and you have

    [tex]\sin(2 \theta)/2 = \frac{\Delta x a}{v_0^2}[/tex]
    which you can easily solve for the angle.

    For the second one, I guess you mean an axis passing through the middle of a ring, perpendicular to the surface of ring. The force for it is:
    [tex]\frac{kqQx}{(x^2 + a^2)^{3/2}}[/tex]
    where a is the radius of circle, Q is the charge of ring, and q is our hero. It appears that you've assumed a>>r, so the force would be simplified into
    [tex]\frac{kqQx}{a^3}[/tex]. Plugging this to 2nd law of Newton
    [tex]\frac{md^2}{dt^2} = \frac{kqQx}{a^3}[/tex]
    It's solution Asin(wt) + Bcos(wt) where w is [tex]\sqrt{\frac{kqQx}{ma^3}}[/tex]. Since [tex]\omega = 2 \pi f[/tex]...

    And finally, or the third: E at (0,y) should be the superposition of all three. For q charges, it's [tex](kq/{r^2}) \sin(\theta) = (kq/{r^2}) \frac{y}{r}[/tex], and [tex]r=\sqrt{y^2+a^2}[/tex]. There're 2 of them. And for -2q, it's simply [tex]k(-2q)/{y^2}[/tex]. Just add the 3 then.
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