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What is the magnitude & direction of the eletric field?

  1. Apr 16, 2017 #1
    1. The problem statement, all variables and given/known data
    Two +3.0nC charges are shown in the diagram below which are spaced 10cm apart. What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

    What is the electric field in N/C?
    What is the direction of the electric field? (above the horizontal)

    2. Relevant equations
    E = kq/r^2

    3. The attempt at a solution

    For the top charge:
    E1 = (8.99*10^9)(3*10^-9C)/(.05m)^2 = 10788 N/C
    E2 = (8.99*10^9)(3*10-9C)/(.1118m)^2 = 2157.7 N/C

    Add them together and I got 1.3 * 10 ^4 N/C

    My answer was wrong... was it because the two charge have the same signed charge so really the y-components cancel so the magnitude of the electric field is just 2*E1?
     

    Attached Files:

  2. jcsd
  3. Apr 16, 2017 #2

    gneill

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    Staff: Mentor

    The electric field is a vector quantity. You need to be adding vectors, not magnitudes.
     
  4. Apr 16, 2017 #3

    Doc Al

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    Staff: Mentor

    The field contribution from each charge is a vector. You must add them as vectors, not just numbers. Direction counts!
     
  5. Apr 16, 2017 #4
    Oh, okay so for the E1 (Top charge):

    E1 = ((8.99*10^9)(3*10-9)/(.05^2) i, 0j)

    E2 (Bottom charge):
    E2 = ((8.99*10^9)(3*10-9)/(.05^2) i, (8.99*10^9)(3*10-9)/(.1^2) j)

    Etotal = (21576 i, 2697j)

    |Etotal| = 21743.909 N/C ? I already tried this and they said it was wrong
     
  6. Apr 16, 2017 #5
    ^^ see reply. I am a little confused about the directions.
     
  7. Apr 16, 2017 #6

    gneill

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    This calculation is not correct. You can't take the components of the distance vector and use them individually as distances for the calculation of the field components. You need to use the scalar distance between the charge and the location of interest in the Coulomb equation when finding the magnitude of the field, then break that magnitude into x and y components according to the vector direction. The radial distances are ##r_1## and ##r_2## in the figure below:

    upload_2017-4-16_17-49-35.png
     
  8. Apr 16, 2017 #7
    Okay so I am a little confused but for Q1 the electric field would be:
    10788 N/C...

    for Q2 it would be 2157.73 N/C...

    The components of Q1 would be (10788i, 0j) & for Q2 it would be (2157.7cos(60) i, 2157.7sin(60) j) ?
     
  9. Apr 16, 2017 #8
    & The magnitude of the field would be 1.2*10^4 N/C & the direction would be 8.95 degrees above the horizontal?
     
  10. Apr 16, 2017 #9

    gneill

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    Close, but is the angle really exactly 60 degrees? How did you determine the angle?

    Note that when you have similar triangles (in the geometric sense of "similar"), it's often easier to just work with the ratios of known side lengths to form the trig functions. So, for example, the cosine of your angle might be formed as:

    ##cos(θ) = \frac{5}{\sqrt{5^2 + 10^2}}##
     
  11. Apr 16, 2017 #10
    ohhh! thank you so much!

    the direction of the field will be 9.32 degrees above the horizontal
     
  12. Apr 16, 2017 #11

    gneill

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    Staff: Mentor

    Yes! :approve:
     
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