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Physics problems.

  1. Dec 19, 2003 #1
    Problem 12a.
    Given:k_e=8.98755*10^9Nm^2/C^2.
    A 5u_C point charge is on the x-axis at x=2.3m, and a 4u_C point charge is on the x-axis at x=3.3m.
    Determine the magnitude of the net electric field at the point on the y-axis where y=2.1 m. In N/C.
    Note: I been getting wrong answers from a problem that is bascially the same as problem 12a. Can someone show me the steps?
     
  2. jcsd
  3. Dec 20, 2003 #2

    ShawnD

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    I think I know where you went wrong.
    The charges are a different distance and angle from the point on the Y axis. This means that the fields are not in phase with each other. You have to sum the fields vectorally.


    If you still can't get the answer, post your work.
     
    Last edited: Dec 20, 2003
  4. Dec 20, 2003 #3
    I still don't get it. Here's my work:
    q_1=5*10^-6
    q_2=4*10^-6
    y=2.1 m

    Since 5u_C is on the x-axis with 4u_C the distance between them was 1m.
    So I found my angle by tan-1(2.1/1)=64.53665494.

    E_1=(8.99*10^9)(5*10^-6)/(2.1^2)=10192.74376
    E_2=(8.99*10^9)(4*10^-6)/(2.32594067^2)=6646.950092

    E_x,2 =(E_2)*cos64.53665494=2857.747052

    E_y,2=-(E_2)*sin64.53665494=-6001.268809

    E,x tot=2857.747052
    E,y tot = 4191.474951
    Etot=square root(2857.747052)^2+((4191.474951)^2=5072.985362N/C

    55.74383949 degrees.
     
  5. Dec 20, 2003 #4

    ShawnD

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    Ok I sort of understood what you were doing but not completely. I'll try to get the answer here.

    field from the first charge:

    [tex]E = \frac{kq}{d^2}[/tex]

    [tex]E = \frac{(9x10^9)(5x10^-^6)}{\sqrt {2.3^2 + 2.1^2}}[/tex]

    [tex]E = \frac{45000}{3.114}[/tex]

    [tex]E = 14451[/tex]


    field from second charge:

    [tex]E = \frac{kq}{d^2}[/tex]

    [tex]E = \frac{(9x10^9)(4x10^-^6)}{\sqrt{3.3^2 + 2.1^2}}[/tex]

    [tex]E = \frac{36000}{3.9115}[/tex]

    [tex]E = 9203.6[/tex]



    Now to solve for the resultant I drew a triangle.
    http://myfiles.dyndns.org/pictures/triangle.jpg

    Where the 2 vectors meet, I drew a line left and one down to help show the angle for that intersection. Now here are the angles for each of the labels

    angle A:

    [tex]A = tan^-^1(\frac{2.1}{3.3})[/tex]

    [tex]A = 32.47[/tex]


    angle B is just 90

    [tex]B = 90[/tex]


    angle C:

    [tex]C = tan^-^1(\frac{2.3}{2.1})[/tex]

    [tex]C = 47.6[/tex]

    The angle where the vectors meet is just the sum of those angles.
    so the angle is 32.47 + 90 + 47.6 = 170 degrees


    Now use the cosine law to find the resultant

    [tex]C = \sqrt{A^2 + B^2 - 2ABcos(c)}[/tex]

    [tex]C = \sqrt{14451^2 + 9203.6^2 - 2(14451)(9203.6)cos(170)}[/tex]

    [tex]C = 23569[/tex]


    And that's what I think the answer is.
     
  6. Dec 20, 2003 #5
    So the magnitude of the net electric field at the point on the y-axis where y=2.1m is 23569 and the angle is 170.
     
  7. Dec 20, 2003 #6
    6.6187*10^3 N/C

    This is what I got when I retried it again. Is this right?
     
  8. Dec 20, 2003 #7

    ShawnD

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    The angle for the resultant is not 170. 170 is the angle between the electric fields. You don't need to know the angle of the resultant, the question does not ask for it.
     
  9. Dec 20, 2003 #8
    I'm just wondering what the angle of this electric field? Would I divide 14451 by 9203.6 and multiply that by tan-1 to get the angle.
     
  10. Dec 20, 2003 #9

    ShawnD

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    If you wanted the angle of the resultant, you wouldn't do the vector triangle method like I did. What I probably would have done is find the angle for each field then break each field into x and y components. Add the x components together, add the y components together then solve for the resultant like it's a right angle triangle. From there, since you know the x and y components, you could just use tan to find the angle.
     
  11. Dec 20, 2003 #10
    6.6187*10^3 N/C

    This is what I got when I retried it again. Is this right?
     
  12. Dec 20, 2003 #11

    ShawnD

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    My answer was 23569; that doesn't mean I'm right though.

    In the first post you said
    and that this was problem 12a.
    Can you post the other problem as well as the answer?
     
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