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E-field, force, & changing sign of charge (grade 12)

  1. Feb 16, 2016 #1
    1. The problem statement, all variables and given/known data
    a) Find the net electric field at a point
    b) Find the net force experienced by a charge of -1.2 x 10-5 C placed at that point
    c) What would happen if the charge changed signs?

    2. Relevant equations
    FE = qε

    3. The attempt at a solution
    a) Solved: 6.7 x 104 N/C [E 63° N]

    b) FE = (6.7 x 104 N/C)(1.2 x 10-5 C)
    = 0.80 N [S 27° W]

    I'm not sure about this. I think I'm only supposed to take the magnitude of the charge into account, and figure out directions after? Also am I correct to use the net field I've already calculated?

    c) The magnitude of the force would remain the same, but the directions would swap since the positive charge is being attracted rather than repelled (0.80 N [E 63° N]).
     
  2. jcsd
  3. Feb 16, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    You can do that, or you could separate the field vector into individual components (East and North, corresponding to X and Y components in a Cartesian coordinate system), then compute the force components individually, recombining the resulting components into your "geographical polar" form afterwards.

    Presumably so, although without seeing the rest of the problem statement we can only assume that your field is correct.

    If by "swap" you mean a 180° direction change, then yes.

    You probably don't want to use "attracted rather than repelled" as a descriptor since that implies an object to which something is attracted or repelled. Here, as far as we can tell from the information you've supplied, there's just a field in space without a source being defined.
     
  4. Feb 16, 2016 #3
    Thanks for your response. I'll add the missing info.

    a) Calculate the net electric field at point A for each of the following charge distributions.

    IMG_3288.JPG

    Ex = (9.0 x 109 Nm2/C2)(3.0 x 10-5 C)/(3.0m)2
    = 3.0 x 104 N/C [East]

    Ey = (9.0 x 109 Nm2/C2)(6.0 x 10-5 C)/(3.0m)2
    = 6.0 x 104 N/C [North]

    Enet2 = (3.0 x 104)2 + (6.0 x 104)2
    Enet = 6.7 x 104 N/C

    θ = tan-1(6.0 x 104/3.0 x 104)
    = 63°
    ∴ Enet = 6.7 x 104 N/C [E 63° N]


    Alright, I think I understand.

    c) The direction of the force experience by the particle at point A would change directions because the negative charge to the east is now attracting rather than repelling the charge, as is the negative charge to the north of point A. The new net force is therefore 0.80 N [E 63° N].
     
  5. Feb 16, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    That looks good.
     
  6. Feb 16, 2016 #5
    Thank you
     
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