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Homework Help: Electric field strength at the midpoint between the two charges.

  1. Jun 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A + 7.3 nC point charge and a - 2.4 nC point charge are 3.5 cm apart. What is the electric field strength at the midpoint between the two charges?


    2. Relevant equations
    E = 9*10^9(q/r^2)
    q = charge
    r = distance from point charge


    3. The attempt at a solution

    Since the question asks for the field strength between the two charges, r would be 1.75 cm or .0175 m.
    Therefore

    E = E1+E2
    E1=9*10^9(7.3*10^-9/.0175^2)
    E1=214531
    E2=9*10^9(-2.4*10^-9/.0175^2)
    E2=-70530.6
    E=144000 N/C

    Except it says I'm wrong. Its multiple choice, and the other choices given are
    1900 N/C
    5700 N/C
    2.8×10^5 N/C
     
  2. jcsd
  3. Jun 5, 2010 #2

    rl.bhat

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    Homework Helper

    If you keep a positive test charge at the mid point, positive charge will repel it and negative charge will attract it. So E1 and E2 are in the same direction. Hence

    E = E1+E2 = 214531 + 70530.6 = ...?
     
  4. Jun 5, 2010 #3
    So, to make this work, would my E2 equation have to be E=9*10^9(q/-r^2)? I don't know what you mean when you say E1 and E2 are in the same direction.
     
  5. Jun 5, 2010 #4

    rl.bhat

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    Homework Helper

    The magnitude of E2 = k*q2/r2 and the direction is towards -2.4 nC.

    The magnitude of E1 = k*q1/r2 and the direction is towards -2.4 nC.

    where r = d/2.
     
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