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Force and Torque on a dipole

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A point charge Q1 is distance r from the center of a dipole consisting of charges +/- q2 separated by distance s. The charge is located in the plane that bisects the dipole. At this instant, assuming r>> s,

    Part A: what is the magnitude of the force on the dipole?
    Part B: what is the magnitude of the torque on the dipole?

    In the space provided, enter the factor that multiplies (1/ε0) in your answer. Express this factor in terms of q2, Q1, s, π, and r.

    2. Relevant equations
    1. Edipole=(1/4πε0)(2p/r3)
    2. p=qs
    3. F=qE

    4. τ=pEsinϑ

    3. The attempt at a solution

    Part A
    Substituting equation 1and 2 into equation 3
    F = qE
    = (q2)(1/4π)(2q2s/r3)

    Part B
    τ = pEsinϑ
    = (q2s)(1/4π)(2q2s/r3)sin90º
    = (q2s)(1/4π)(2q2s/r3)


    I got a feeling I'm missing an important step involving Q1 in both parts.
    Help's much appreciated =)
     
  2. jcsd
  3. Jan 31, 2009 #2

    rl.bhat

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    Homework Helper

    Since r>>>>s force on +q and -q are equal and opposite. So what will be the net force on the dipole?
    Torque = forceX distance.
    Find the force between Q1 and q and calculate the torque.
     
  4. Jan 31, 2009 #3
    thanks for the reply rl.bhat
    but for Part A, the question is looking for the magnitude therefore the force on one end of the dipole and not the net force. the net force on a dipole would be zero of course.

    As for part b, i doubt that the solution can be that simple. The question must be solved through algebraic expressions only. Solving it in this manner would require an extensive amount of substitution.

    If solved simply by using the forces between Q1 and +/-q2, we would have to add their adjacent component values requiring inverse tangent and cosine laws
     
  5. Jan 31, 2009 #4

    rl.bhat

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    Homework Helper

    The expression which you have given for torque is for a dipole kept in a uniform electric field. Since r >>> s , you can assume that the field is uniform.
    In your solution of part B, you have to substitute the field due to charge Q1 at q2.
     
  6. Jan 31, 2009 #5
    1 gives the radial dipole field at a point on the line of the dipole charges

    At points on the "bisecting plane" the field due to the dipole is "tangential"
    Edipole=(1/4πε0)(p/r3)
    (Although the point of the problem is to derive this)
    This field exerts a force on the point charge,
    Q1(1/4πε0)(p/r3)

    and the point charge exerts an equal and opposite force on the dipole.

    Note the moment of this couple is equal and opposite to the torque,
    necessary for conservation of angular momentum.
     
    Last edited: Jan 31, 2009
  7. Feb 1, 2009 #6
    if understand part b properly
    t = force x distance
    where force is the line action, therefore s sin ϑ which is basically s
    force is the force between q2 and q1
    so overall it would be
    Q1(1/4πε0)(q2s/r3)s
     
  8. Feb 1, 2009 #7
    For ease of reading/writing, I'll call the point charge Q and the dipole charges+/- q.

    The magnitude of the force exerted by Q on +/-q is F = Qq/(4 pi epsilon0 r^2)

    a) Add the forces vectorially (triangle of forces) to find the resultant R.
    Note that the triangle is similar to the one with sides r,r and s:
    s/r = R/F

    b) F*s
     
  9. Feb 1, 2009 #8
    awsome i got it
    big thanks davieddy!
     
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