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Gravity + Electric field

  1. Apr 13, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    On a certain planet, the acceleration dute to gravity is the same as that on earth but there is a strong downward electrical field with the field being uniform close to the planet's surface. The magnitude of this electric field is E= 4.75x10^5N/C. A 1.00kg mass carrying a charge of [tex]100\mu C[/tex] is connected to a vertical spring for which the spring constant is 200N/m. The spring is not stretched when the mass is released from rest.

    a) by what maximum ammount does the spring expand?
    b) What is the equillibrium postition of the mass
    c) show that the mass executes simple harmonic motion

    2. Relevant equations

    3. The attempt at a solution

    [tex]E= 4.75x10^5 N/C[/tex]
    [tex]m= 100kg[/tex]
    [tex]q= 100\muC [/tex]
    [tex]k= 200N/m[/tex]
    [tex]g= 9.8m/s^2 [/tex]

    hm..I would think that the downward gravitational field would make the force pushing a person or mass down would be greater.

    sort of lost here.

    Can someone help? Thanks

    I was thinking that well there is force downward on the spring that is initially unstretched but this Fy is composed of electric field and gravity so:

    E= F/q
    I think I can find the force since the E is given and the q is given right?

    Then would I just add that to the gravitational force?

    and then use that on spring as in F=-kx since I would have the force ?
     
    Last edited: Apr 13, 2008
  2. jcsd
  3. Apr 13, 2008 #2

    Nabeshin

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    What's the issue? You have all your steps laid out, why don't you do it? Have a little faith in yourself, especially when you have no problems in your logic.

    P.S: q is not 200. Might just be a mis-type but that would be one huge charge ^^
     
  4. Apr 13, 2008 #3

    ~christina~

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    I said k was 200...never said q was 200.

    see the thing is that it's not really the first step that is the problem usually it's the 2nd and and 3rd step that have me like this :uhh:
     
  5. Apr 13, 2008 #4

    Nabeshin

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    Mistake, I meant 100. In your "attempt at solution" it says q=100, which should be 100*10^-6. Oh and you wrote m=100kg, which should be 1kg lol.
     
  6. Apr 13, 2008 #5

    ~christina~

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    [tex]q= 100x10^-6 C [/tex]

    a and

    [tex]m= 1.00kg[/tex]

    alright so...if I did the problem like I said I would then I would get:

    E=F/q

    F=Eq

    F= (4.75x10^5 N/C) (1.00x10^-6C)= .475N

    and the force due to gravity equals

    F=mg

    F= 1.00kg(9.8m/s^2)= 9.8N

    [tex]F_{total}= .475N + 9.8N= 10.275N[/tex]

    and that would be put into

    F=-kx

    10.275 N/(200N/m)= x

    x=5.1375x10^-2 m

    but as to finding 2 is it using the fact that when a object is at equillibrium that v=0?

    Thanks
     
  7. Apr 13, 2008 #6

    Nabeshin

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    Actually, that's the equilibrium position you found. Equilibrium is defined as when the sum of the forces and torques is equal to zero. However, because of good old newton's first, the object will continue moving through this point. Consider the work the forces (downward, constant forces) and spring (changing force) will do and the point at which v=0 is the maximum displacement.
     
  8. Apr 13, 2008 #7

    Doc Al

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    Check that value for charge.

    OK.
    As Nabeshin points out, here you have calculated the equilibrium position (measured from the starting point), which is the point where the net force is zero.

    Use energy conservation to solve for the lowest point--that's the point where v = 0.
     
  9. Apr 13, 2008 #8

    ~christina~

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    fixed

    [tex]q= 100x10^-6 C [/tex]

    a and

    [tex]m= 1.00kg[/tex]

    alright so...if I did the problem like I said I would then I would get:

    E=F/q

    F=Eq

    F= (4.75x10^5 N/C) (100x10^-6C)= 47.5N

    and the force due to gravity equals

    F=mg

    F= 1.00kg(9.8m/s^2)= 9.8N

    [tex]F_{total}= 47.5N + 9.8N= 57.3[/tex]

    and that would be put into

    F=-kx

    57.3N/(200N/m)= x

    x=2.865x10^-1 m

    __________________
    a)

    for the conservation of energy I'd say that...at max displacement
    PE= 1/2kA^2
    KE= 0
    x= A

    initial it's

    [tex]PE_i + KE_i + PE_si= KE_f + PE_f + PE_sf[/tex]

    [tex]mgh + 0 + 0= 1/2mv^2 + 0 + 1/2kx^2[/tex]

    right or not?
     
  10. Apr 13, 2008 #9
    Don't forget electric potential energy.
     
  11. Apr 13, 2008 #10

    ~christina~

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    [tex]PE_i + KE_i + PE_si+ PE_ei= KE_f + PE_f + PE_sf+ PE_ei[/tex]

    [tex]mgh + 0 + 0 + q_o \Delta V = 1/2mv^2 + 0 + 1/2kx^2[/tex]

    is that right? not sure how the equation for potential Electrical Energy would look if in the conservation of energy equation.

    Thanks
     
    Last edited: Apr 13, 2008
  12. Apr 13, 2008 #11
    The electric field is uniform, but you're using the PE associated with the field of a point mass.
     
  13. Apr 13, 2008 #12

    ~christina~

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    so..is it fine now?

    [tex]mgh + 0 + 0 -q_o Ed = 1/2mv^2 + 0 + 1/2kx^2 + 0[/tex]

    Thanks
     
    Last edited: Apr 13, 2008
  14. Apr 14, 2008 #13

    Doc Al

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    I assume you're solving part a? So you're comparing the energy at the point of release with the energy at the lowest point.

    (1) Why is the electrical PE negative? The electric force points downward, just like gravity.
    (2) Why is there a KE term on the right? At the lowest point, the speed is zero. (If it wasn't zero, that means it would still be moving down and thus not yet at the lowest point.)

    Be sure to express the distances in terms of the same variable. I see h, d, and x: all you need is one.
     
  15. Apr 14, 2008 #14

    ~christina~

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    okay, trying to fix this based on what you said.

    [tex]mgd + q_o Ed + 1/2mv^2 = 1/2kd^2 [/tex]

    about the electric field: how would I determine if it is possitive or negative?
     
  16. Apr 14, 2008 #15

    Doc Al

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    Almost there: It's released from rest.

    You are told that the field is downward and the charge is positive (I assume), so the electric force acts downward just like gravity.
     
  17. Apr 14, 2008 #16

    ~christina~

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    hm...

    [tex]mgd + q_o Ed = 1/2kd^2 [/tex]

    so if the charge is negative, would that just make the sign negative since it would go up instead of downwards?
     
  18. Apr 14, 2008 #17

    ~christina~

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    okay I don't know how to show that something executes simple harmonic motion.

    need help on that last part.

    Thank you
     
  19. Apr 14, 2008 #18

    Doc Al

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    Good. Now you can solve for d.

    Right.

    Show that the restoring force is proportional to displacement from equilibrium. (Like a spring obeying Hooke's law: F = -kx.)
     
  20. Apr 14, 2008 #19

    ~christina~

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    [tex]mgd + q_o Ed = 1/2kd^2 [/tex]

    um...well I seem to have issues with finding the d but I got this far..

    (1.00kg)(9.8m/s^2)d +(100x10^-6C)(4.75x10^5 N/C)d= 1/2(200N/m)(d)^2
    9.8Nd + 47.5Nd= 100N/m d^2

    100N/m d^2 - 57.3d= 0

    (I know I factor but I'm not sure what number to factor with for this...for example if it is
    2x^2-4x= 0 then the equation would be factored to 2x(x-2)= 0 and thus 2x=0 or x-2= 0 and answer would be 2)

    "restoring force"

    wouldn't that be hooke's law, though?
    (didn't really get it in class and can't find it online)

    I'm being told to draw it but for a exam I don't think it would be acceptable to draw.

    Thanks.
     
  21. Apr 14, 2008 #20

    ~christina~

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