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Net Electric Field Strength

  1. Oct 17, 2014 #1
    1. The problem statement, all variables and given/known data

    A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 3600 N/C and is directed in the positive x direction. A point charge -9.8 × 10-9 C is placed at the origin. Find the magnitude of the net electric field at y = +0.20 m.


    2. Relevant equations
    E=kq/r^2

    3. The attempt at a solution
    I thought at first because the the y-direction is perpendicular to the electric field direction that the field strength would be 0, but this is wrong. I then thought the field would be the same at that point i.e. be 3600N/C, but that was wrong too. So how do I correctly solve this problem?
     
  2. jcsd
  3. Oct 17, 2014 #2

    Orodruin

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    Remember that you have two contributions to the electric field here. One from the uniform field and one from the added point charge.
     
  4. Oct 17, 2014 #3
    So because the uniform field acts perpendicular to the field from the point charge, the electric field at y=0.2m would be just the electric field from the point charge alone? Is my understanding correct?
     
  5. Oct 17, 2014 #4

    collinsmark

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    That's not quite right. The electric field at the location y = 0.2 m, is [itex] \mathrm{(3600 \ N/C)} \hat a_x [/itex] plus whatever the electric field is from the point charge. (In my notation, I am using [itex] \hat a_x [/itex] as the unit vector pointing in the x direction.)

    Don't forget that they are vectors. (Don't merely sum the magnitudes.) [Edit: Hint: you need to break up each, individual electric field into its components, then you may sum the corresponding components of the individual fields.]
     
  6. Oct 17, 2014 #5
    So 3600N/C field acts in the x-axis direction, and the field from the point charge acts in all directions with a value of (9x10^9 x 9.8 x 10^-9)/0.2^2. So what I do is add these two together to find the total electric field in the x axis direction (let this sum be A), and the field in the y axis direction is just the field from the point charge (let this sum be B). And then I can use Pythagoras' formula here to say total electric field at y=0.2m = square root of (A^2+B^2). Is this now correct?
     
  7. Oct 18, 2014 #6

    Orodruin

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    No, does the point charge give a field component in the x direction?
     
  8. Oct 18, 2014 #7
    Doesn't the point charge give a field in all directions?
     
  9. Oct 18, 2014 #8

    Orodruin

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    It gives a field pointing in the radial direction. Along the y-axis, the radial direction is the y-axis and therefore the point charge does not give any component in the x-direction.
     
  10. Oct 18, 2014 #9
    Oh I see. So basically letting A= field due to point charge (which is in y axis direction) and B = field due to field in x-axis direction, because they are at right angles to each other net electric field at y=0.2m = square root of [ [(9x10^9 x 9.8x10^-9)/(0.2^2)]^2 + (3600^2)]?
     
  11. Oct 18, 2014 #10

    Orodruin

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    That would be the correct idea, yes.
     
  12. Oct 18, 2014 #11
    The answer still came out wrong?
     
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