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Electric Field at two points

  1. Jul 13, 2017 #1
    1. The problem statement, all variables and given/known data
    A 90 μC point charge is at the origin.
    Find the electric field at the point x3 = -35 cm , y3 = 80 cm .
    Express your answers using two significant figures. Enter the x and y components of the electric field separated by a comma.

    2. Relevant equations
    E = kQ/r^2
    magnitude * cos/sin of theta

    3. The attempt at a solution
    So I found the Etot to be 1.06*106 N/C and from there I drew a picture and found each component by using sin and cos of theta so I get Ex = 1.06*106*cos(.35/.7625) and 1.06*106*cos(.8/.7625) for the Ey. This is wrong however. So what steps am I missing here?
     
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  3. Jul 13, 2017 #2

    gneill

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    How did you arrive at the cos and sin arguments? I don't see where you've accounted for the sign of the x3 value in your component calculations; do you expect the field vector to have two positive components? What does a look at your sketch of the scenario suggest? (you did make a sketch first, right?)
     
  4. Jul 13, 2017 #3
    This is all my work. I found the total E field magnitude, but I need it in components. So originally I just did it in components with each x and y value. This I realized is wrong. Both x and y contribute to the total r in the E field equation for a point charge. I then decided ok, I have magnitude, but need components not net. I then thought that I could simply take the magnitude E and multiply by the corresponding angle using sin or cos. Is sin or cos not even used?
     

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  5. Jul 13, 2017 #4

    haruspex

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    You seem to have a basic misunderstanding of trigonometry.
    The sine and cosine are functions of the angle, but the angle is not the ratio between the two distances.
    In fact, you do not need to find the angle, since you have the distances and the sine and cosine values come directly from their ratios.
    For the x component, the cosine of the angle of r to the x axis is just the x distance divided by r.
    (Where did you get 0.7625 from? r must be more than 0.8.)
     
  6. Jul 13, 2017 #5
    0.7625 is r. I just did (0.352+0.82) and then took the square root of that which gave 0.7625. So its cos(.35/r) for x?
     
  7. Jul 13, 2017 #6

    haruspex

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    No, it looks like you forgot to do that. Look at the triangle. How could r be less than y?
    No. If θ is the angle r makes to the x axis then cos(θ)=.35/r.
     
  8. Jul 13, 2017 #7
    I see so r is 0.8732 and cos(θ)=.35/.8732 = .400819. Then multiply by E to get x-component?
     
  9. Jul 13, 2017 #8

    haruspex

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    Yes.
     
  10. Jul 13, 2017 #9
    Ok so I get for the x-component 4.3*105 N/C and for the y-component I get 9.7*105 N?C. I am asking since I only have one attempt left. I took the 1.06*106 * (0.35/0.8732) for the x-component and 1.06*106 * (0.80/0.8732) to get the two above numbers.
     
  11. Jul 13, 2017 #10

    haruspex

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    That's essentially correct, but I make the x component just under 4.25, so closer to 4.2 than 4.3. Not sure how accurate you need to be. I recommend you run the calculations again keeping one more digit of accuracy until the end. I can believe your 4.3 is right and I am wrong.
     
  12. Jul 14, 2017 #11

    ehild

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    What is the correct sign of Ex?
     
  13. Jul 14, 2017 #12
    It is negative. Thanks for the help everyone!
     
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