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Coulomb's Law Problem magnitude and direction

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data

    What are the magnitude and direction of the electric field at a point midway between a -9.0 µC and a +6.0 µC charge 3.0 cm apart? Assume no other charges are nearby.

    2. Relevant equations

    Fe= {K(q1)(q2)}/d^2

    k = (8.99 x 10^9)

    3. The attempt at a solution

    First I converted the µC to C =
    -9 µC = -.000009 C
    6 µC = .000006

    Then convert 3 cm to mm = .03

    (8.99 x 10^9) (-.000009) ( .000006 ) = -.48546

    -.48546/d^2

    -.48546/ .03

    -539.4

    My final answer is -539.4 but I know its wrong, can anyone tell me what I did incorrectly?
     
  2. jcsd
  3. Apr 1, 2009 #2

    LowlyPion

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    You found the Force between the 2 charges. The question asks for the field at a point half way between the two.

    F = q*E

    What you want is E, given by

    E = kq/r2

    Your distance is .015 m to each. And remember these are vectors and when you add the field at the mid point you want to take care with what direction the field of each would be pointing.
     
  4. Apr 1, 2009 #3
    What Q value would you use though?
     
  5. Apr 1, 2009 #4

    LowlyPion

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    For E you are using the q of the charges you are measuring. 1 term will be the field from the 6 µC charge and the other the 9 µC.
     
  6. Apr 1, 2009 #5
    So there will therefore be two answers?

    I'm confused by what your saying
    "For E you are using the q of the charges you are measuring."
    I realize this, but do I need to set up two equations then? Or do I have to add the charges?
     
  7. Apr 1, 2009 #6

    LowlyPion

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    The E-field at any point will be the ∑ of the E from all charges
     
  8. Apr 1, 2009 #7
    so sigma is the sum right?


    E = k(6x10^-6)/r2
    is added with:
    E = k(-9*10^-6)/r2

    once I've added these two Es up is that my final answer?
     
  9. Apr 1, 2009 #8

    LowlyPion

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    Sort of. But you must be careful with the sign. Since one charge is plus and the other minus, in between the charges then the field magnitudes both add and point toward the negative charge.
     
  10. Apr 1, 2009 #9
    {(8.99*10^9)(-9*10^-6)}/.015^2 = -359,600,000

    {(8.99*10^9)(6*10^-6)}/.015^2 = 239, 733, 333

    Adding these two up

    -359,600,000 + 239, 733, 333 = -119, 866, 666

    -119, 866, 666 N/C seems ridiculous, what did I do here?
     
  11. Apr 2, 2009 #10

    LowlyPion

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    Not quite.

    As I said several times, these are vectors. As it turns out both vectors are pointing in the same direction.

    To determine the |F| then you simply add the |Fq1| + |Fq2|.

    The - sign means that the Force points toward a charge and the + sign means that the field points away from the charge. Draw a picture nd you will see that at the mid point for these charges the E-field from both is pointing in the same direction ... hence the magnitudes add at that point.

    Oh, and your answers have too many significant figures.
     
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