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[Point Charges] Can't figure this out

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data

    In the figure below, six charged particles surround particle 7 at radial distances of either d = 1.0 cm or 2d, as drawn. The charges are q1 = +8e, q2 = +8e, q3 = +e, q4 = +8e, q5 = +8e, q6 = +4e, q7 = +4e, with e = 1.60 10-19 C. What is the magnitude of the net electrostatic force on particle 7?


    ---2----
    ---|-----
    1-7-3-4
    ---|-----
    ---5----
    ---|-----
    ---6----


    2. Relevant equations

    F = k(q1)(q2)/(r^2)
    k = 8.99e9

    3. The attempt at a solution

    I split this up into x and y components, and then figured out the sum.

    For the x, ƩF = |k(q7*q1)/(r^2) - k(q7*q3)/(r^2) - k(q7*q4)/(4(r^2))|
    This simplifies to: |[k(q7)/(r^2)]*[q1-q3-q4/4]|
    and so ƩF = |[4ke/(.01^2)]*e(8-1-8/4)| = 4ke/.0001*5e = 20k(e^2)/.0001

    I used the same process for the y component. 2 and 5 cancel out since they have the same distance and charge, so the only charge I needed to calculate was for 6. F = k(q7)(q6)/(4(r^2))
    This simplifies.. F = k(6e)(4e)/((4)(.0001)) = 6k(e^2)/.0001

    Now the net charge is ([20k(e^2)/.0001]^2 + [6k(e^2)/.0001]^2)^.5
    Simplify this to [436(k^2)(e^4)/(10^-8)]^.5 which comes out to 4.8055e-23, which webassign says is incorrect.
     
  2. jcsd
  3. Aug 30, 2012 #2

    PeterO

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    I would have calculated the force between q7 and q3 [since q7 is the central charge we are analysing and q3 is just a single e. [lets call that F]

    You can then express every other force in terms of that F

    eg q1 is the same distance, but q1 is +8e so the force is 8F
    q4 is the same size as q1, but twice the distance , so 2F [inverse square law]
    etc.
     
  4. Aug 30, 2012 #3
    So, ƩFx = F + 8F + 8F/4 and ƩFy = 4F/4
    ƩF = (11F^2+F2)1/2
    ƩF = (122F2)1/2
    F = k(4e)(e)/(.012) = 4ke2*.012

    ƩF = 1.0168e-30

    This doesn't seem right since it's 7 degrees off my previous calculations.
     
  5. Aug 30, 2012 #4

    PeterO

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    The x-direction 8F force is in the opposite direction to the other two.
     
  6. Aug 30, 2012 #5
    Ahh... so ƩF = (5F2+F2)1/2 = (26F2)1/2

    This still results in a suspiciously large exponent: ƩF = 4.694e-31
     
  7. Aug 30, 2012 #6

    PeterO

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    Is the "real" answer of the order of 10-22 by any chance?
     
  8. Aug 30, 2012 #7

    PeterO

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    Looks like a divide changed to a multiply ??
     
  9. Aug 30, 2012 #8
    Wow. All I can say is, stay off drugs, kids.

    edit: So I ended up with a value of 1.016e-22, and it's STILL not right. I've run out of chances now, but I'm still interested in knowing what I'm doing wrong here.
     
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