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Electric dipole in uniform electric field

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data
    The prototypical electric dipole is made of two equal and opposite charges of magnitude q separated by a constant distance d (for example, the charges can be placed at opposite ends of a length d rod).
    The dipole is placed in a uniform, external electric field Er = E0 yˆ . The positive charge is at coordinates (x, y, z) = (+d/2 sinθ, +d/2 cosθ, 0) and the negative charge is at coordinates (x, y, z) = (–d/2 sinθ, –d/2 cosθ, 0).
    !
    a) What is the electric force on each charge? What is the net force on the dipole?
    b) What is the net torque on the dipole? At what angle(s) will the net torque be zero?
    c) The dipole is initially oriented perpendicular to the electric field (θ = 90°). How much work does it take to rotate the dipole so that it is parallel to the electric field (θ = 0°)? Anti- parallel (θ = 180°)? (Hint: Take the voltage to be zero at the origin and find what potential difference each charge moves through.)
    d) Now consider the dipole oriented along the x-axis and centered on the origin. Sketch this configuration, and draw some field lines (using the convention that a charge q gets 8 field lines) and at least 5 equipotential lines.


    2. Relevant equations
    F = kQQ/r^2
    p= qd
    t = pE

    3. The attempt at a solution
    F on q+ = (9x10^9)Q^2 / d^2 sin^2 θ+ d^2 cos^2 θ
    F on q- = (9x10^9)Q^2 / d^2 sin^2 θ+ d^2 cos^2 θ
    F net = 0

    τ = Qd^2sin^2 θ + d^2cos^2 θ + E0y^

    I dont know if I am doing this right. Any help would be greatly appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 27, 2012 #2

    SammyS

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    Hello svbalasub. Welcome to PF !

    You have: "F on q+ = (9x10^9)Q^2 / d^2 sin^2 θ+ d^2 cos^2 θ". If Q = |q|, then this is the magnitude of the force each of the charges making up the dipole exerts on the other.

    You need to use [itex]\vec{F}=Q\vec{E}\,,[/itex] where Q = q or Q = -q, depending upon which charge you're working with.
     
  4. Feb 27, 2012 #3
    so it would F = QE0y^ ?
     
  5. Feb 27, 2012 #4

    SammyS

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    What is the "it" you refer to ?
     
  6. Feb 28, 2012 #5
    So is the answer for electric force on each charge just F= KQ^2/d^2??

    Since: F=Kq1q2/r^2
    r= sqrt d^2sinθ+d^2 cos^2 θ =d
     
  7. Feb 28, 2012 #6

    SammyS

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    Hello HelPhysics. Welcome to PF !

    It is true that r = d. Aside from that, the rest of what you have is the wrong answer for what's being asked.

    They're asking for the force on a charge of q, and then a charge of -q, due to an electric field of [itex]\vec{E}=E_0\,\hat{\textbf{j}}\,.[/itex]
     
  8. Feb 28, 2012 #7
    Got it! SO it's just F on -q= -qE0jˆ and F on q= qE0jˆ correct?

    What about part b, the net torque on the dipole? I know we use the equation: T=ƩrxF. Do you have any suggestions how to get started?
     
  9. Feb 28, 2012 #8

    SammyS

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    Yes. Use that equation.
     
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