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Gauss Law and 2 charges

  1. May 27, 2008 #1
    1. The problem statement, all variables and given/known data

    A Particle of mass m and chargeq oves at high speed along the x axis. It is initially near x=-infinity and it ends up near x=+infinity. A second charge Q is fixed at the point x=0, y=-d. As the moving charge passes the stationary chrge, its x component o velocity does not change appreciably, but it acquires a small velocity in the y direction. Determine the angle angle through which the moving charge is deflected. suggestion: The integral you encounter in determining v_y can be evaluated by applying Gauss's law to a long cylinder of radius D, centred on the stationary charge.

    2. Relevant equations

    [tex]\Phi[/tex] = EdA

    3. The attempt at a solution

    So Far, I have managed to get to E=Q/2[tex]\pi[/tex]md([tex]\epsilon[/tex]_0)
    and ended up with a= Qq/(2[tex]\pi[/tex]md([tex]\epsilon[/tex]_0)
    by using electric force and F=ma equations

    I know the answer I should get is [tex]\Theta[/tex]=Qq/(2[tex]\pi[/tex]md([tex]\epsilon[/tex]_0)v[tex]^{2}[/tex]

    can anybody help me with the method between these two points, or correct me if my method is wrong?
  2. jcsd
  3. May 27, 2008 #2
    you are along the right lines here, but you need to consider that the particle will always experience a force, for all x, not just at x = 0, so work out the distance that the particle is for any x, and use this as your distance.

    other than that, dont forget that

    F = q_1 q_2 / 4 \pi \epsilon_0 r^2

    is the force, and from this, and what i said above, you should be able to follow the algebra through and get the right answer =]
  4. May 27, 2008 #3
    OK, so I'm assuming that I am using [tex]r^2=x^2+d^2[/tex], and that [tex]x=r tan \theta[/tex], which should give me the distance, as I can't see another way of getting it.
    but this complicates things on the rhs giving me a "+1 " when simplified, which looks tough to get rid of.

    I still can't see a way of getting the [tex] v^2 [/tex] unless I use kinetic energy, but then that seems to cause more problems than it solves.

    is it the K.E. I'm looking at, or am I missing something a bit more simple?
  5. May 28, 2008 #4
    Yes for the first thing, no for the second. The direction r is the direction of the force on the particle, not the direction of the particle itself.

    first you want to calculate the velocity in the y direction by integrating the force with respect to time, which is the momentum, and hence you have the velocity.

    The angle is calculated from the ratio of the y-velocity to the x-velocity (this can be seen by drawing a triangle, using distance travelled in the x direction and distance travelled in y direction in a given time t, The angle [tex] \theta[/tex] is enclosed by that, and from there is simple substitution).
  6. Jun 13, 2008 #5
    dude don't you get it? there no Vx, so the problem can't be solved. They should give you Vx. It doesnt matter if you use Vy/Vx=tan(()) or Position Y/x=tan(()) you still need the x component.
  7. Jun 13, 2008 #6
    motherfucker then do it.
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