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Electric Field with three charges

  1. Feb 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider 3 charges along a horizontal line as shown:
    8.72e-6 C (+) ___4.16cm____2.45e-6 C (+)______3.18cm_____-1.46e-6C (-)
    The acceleration due to gravity is 9.8 m/s^2
    The Coulomb Constant is 8.98755e 9 Nm^2/ C^2
    What is the electric field at a point 2.18 cm to the left of the middle charge? Answer in units fo N/C


    2. Relevant equations
    Q1= 8.72e-6 C , r= .0198m (.0416m-.0218m)
    Q2= 2.45e-6 C, r= .0218m

    E1= k Q1 / r^2
    E2= k Q2/ r^2

    3. The attempt at a solution
    I converted all of my units first to C and meters.
    Then I solved for E1 and E2 separately.
    Once I had E1 and E2, I added the two together. But I can't seem to get the right answer.
    Am I supposed to subtract them instead becasue Q1 and Q2 are both positive and therefore will repel each other?
     
  2. jcsd
  3. Feb 23, 2009 #2

    Delphi51

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    Homework Helper

    You have THREE charges causing electric fields, so you need to find E1, E2 and E3.
    Add them together, keeping direction in mind!
     
  4. Feb 23, 2009 #3
    so I calculated E1 = 1.999e8, E2 as 4.633e5, and E3 as -3.657e6
    Am I correct to say then that, E1 and E3 are negative (pointing to the left) and E2 is positive?
    E1(+)<----X----->E2 (+)-----><------E3 (-)
     
  5. Feb 23, 2009 #4

    Delphi51

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    E1 to the right (away from a positive charge), E2 to the left (away from a positive charge), E3 to the right (toward a negative charge).
     
  6. Feb 24, 2009 #5
    Thank you!!!!
     
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