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Hai! Calculating Net Electrostatic Force! Need halp :(

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Three charged objects are placed as shown. Find the net force on the object with the charge of -5uC
    Diagram:

    http://imgur.com/xHGHGbd

    2. Relevant equations
    No idea what to do as far as steps go, I don't know how to start :(



    3. The attempt at a solution
    None so far.
     
  2. jcsd
  3. Jan 10, 2015 #2
    Try calculating the magnitude of the forces (individually) acting on the specified object and determine their nature (i.e. attractive/repulsive) , then find the resultant vector of force using trigonometry.
     
  4. Jan 10, 2015 #3
    Okay, I'll go find each one acting on it, gimme a sec :D
     
  5. Jan 10, 2015 #4
    6uc vs. -5uc = -0.0902135052958669(Attractive)
    -7uc vs. -5uc = 0.07875 N(Repulsive)

    How do I use these in trigonometry?
     
  6. Jan 10, 2015 #5
    You've made a unit conversion error somewhere - it should be 9.023...N and 7.875 N.
     
  7. Jan 10, 2015 #6
    Oh, sorry xD
    I might've typed it wrong,or I'm just kinda dumb.
     
  8. Jan 10, 2015 #7
    The forces will be vectors, so construct a vector triangle and find the "length of the missing side" in terms of Newtons.
     
  9. Jan 10, 2015 #8
    Not true - even PhD students can make conversion errors.
     
  10. Jan 10, 2015 #9
    I got 34.6456i, 36.68 deg. S of E, did I get it? :D
     
  11. Jan 10, 2015 #10
    I'm afraid that is incorrect; how can the force have a magnitude which is non-real in this case? I suggest that you use the cosine rule to find the resultant force.
     
  12. Jan 10, 2015 #11
    I did these, what did I do wrong?
    Multiply it by sin(30) and multiply 7.875 again by cos(30) = (6.82 N, 3.94 N)
    Turn the 6uc into vector components. = (-9.02 N,0)
    Add the two vectors = (-2.2, 3.94)
    Use pythagorean theorem on it = 3.27 N
    Arctangent to find direction= -60.809688481761 degrees
    3.37 N, 60 degrees N of W
     
  13. Jan 11, 2015 #12
    The vector triangle will not be right angled - do you have a justification for assuming one of the angles between the vectors to be 90 degrees? The resultant won't be on the same line as the 10cm one, so using pythagoras' principle won't work. Try using the cosine rule.
     
  14. Jan 11, 2015 #13
    4.68 N, 57.4deg S of E
    Got that after various tries, done?
     
  15. Jan 11, 2015 #14
    Multiply it by sin(30) and multiply 7.88 again by cos(30) = (6.82 N, -3.94 N)
    Turn the 6uc into vector components. = (-9.34 N,0)
    Add the two vectors = (-2.52, 3.94)
    Use pythagorean theorem on it = 4.68 N
    Arctangent to find direction= -57.4 Degrees
    4.68 N, 57.4deg S of E
     
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