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Motion of a particle in eletric field

  1. Jan 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys. Just wondering if anyone could help me with Q7 part a) on the pdf I have attached below. I think my working are not correct but I cant see other way. If I add 6. and 7. together I would get an ans in terms of what the question is asking but I am not sure if that is correct?

    upload_2017-1-24_18-19-4.png
    [ note: cropped image inserted by moderator ]

    2. Relevant equations
    1.$$y=y_0+v_{0y}(t)+\frac{1}{2}at^2$$
    2.$$E_y=\frac{F_y}{q}$$
    3. $$F_y=-mg$$
    3. The attempt at a solution

    4.$$y_0=0, v_y=0 ,\uparrow=+, v_{x0}$$
    5.$$E_y=\frac{-mg}{q}, g=\frac{E_yq}{m}$$

    So by subbing in the 4 and 5 values into 1 I get two equations as shown below:
    6.$$y=-\frac{1}{2}g(\frac{d}{v_{x0}})^2$$
    and

    7.
    $$y=-\frac{1}{2}(\frac{E_yq}{m})(\frac{d}{v_{x0}})^2$$


     

    Attached Files:

  2. jcsd
  3. Jan 24, 2017 #2

    Simon Bridge

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    #3 only follows from #2 in a specific situation - does that situation apply for 7a?

    You should start from a free body diagram.
    Use: ##\sum \vec F = m\vec a## ... divide into components and use suvat equations. Y
    ou've done something similar when you learned about ballistics.
     
  4. Jan 24, 2017 #3
    I am slight lost to what you mean. I have used suvat, in the x and y. Could you please expand on this. From the picture the only accleration acting is in the horizontal no acceleration in the vertical. So the horizontal equation looks like so:

    $$d=v_xt$$
     
    Last edited: Jan 24, 2017
  5. Jan 24, 2017 #4

    haruspex

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    You meant the other way around, yes?
    What forces act in the vertical? What is the net vertical force? What vertical acceleration results?
     
  6. Jan 25, 2017 #5

    Simon Bridge

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    Identify the vertical forces and write out Newton's second law accordingly.
    So, in the horizontal direction, Newton's law says: ##\sum F_x = 0 = ma_x: v_x(0)=u, x(0)=0##
    ... when you solve that, you get ##a_x=0, v_x(t)=u, \implies x(t)=ut##

    Good ... so what are the forces in the vertical direction?
    ##\sum F_y = ? = ma_y : v_y(0)=?, y(0)=?## ... fill in the question marks.
     
  7. Jan 25, 2017 #6
    Ok so the net force that act of the particle is:
    $$F_t=E_yq-g=ma_y$$

    using the equation for height:

    $$y=v_{y0}t+\frac{1}{2}a_yt^2$$

    now at $$t=0, v_{y0}=0$$

    therfore my equation becomes:

    $$y=\frac{1}{2}at^2$$

    subbing in my equations for $$a_y ,t$$

    I get:

    $$y=(\frac{1}{2})(\frac{E_yq-g}{m})(\frac{d^2}{v_x^2})$$
     
  8. Jan 25, 2017 #7

    haruspex

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    Looks right.
     
  9. Jan 25, 2017 #8
    Just like to say thank for the guidance.
     
  10. Jan 25, 2017 #9

    Simon Bridge

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    Force of gravity is usually written ##mg##.

    Note: it is usually best practise to use the variable names in the question.
    ##d = uT## and ##2h = aT^2## where T is time between plates... maybe?
     
  11. Jan 25, 2017 #10
    yes sorry, I notice that when i looked back. But i could not edit it.
     
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