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Force on an Electric Dipole

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data
    I got this problem from my physics textbook, and I'm not sure how to solve it.

    A point charge R is distance p from the center of a diple consisting of charges + and - q separated by distance s.

    The charge R is located in the plane that bisects the dipole.

    What is the force (don't forget direction) on the dipole? Assume that p is much bigger than s.


    2. Relevant equations
    Electric field due to a dipole: K2qs / r3, where s is the length of the dipole, q is the charge of the dipole, and r is distance from it.



    3. The attempt at a solution
    The first thing I'm not sure about is the part that says "located in the plane that bisects the dipole". What does that mean?

    I have decided to calculate the force of the dipole on the charge R, rather than the other way around.

    So, the electric field due to the dipole is:
    E = (K)(2qs) / (r3). Multiply this by R to get force.

    However, I don't know how you figure out the direction! Next I have to figure out the torque on this so I will need to calculate the angle the dipole makes with the electric field.

    Ideas?
     
  2. jcsd
  3. Feb 5, 2009 #2

    Delphi51

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    Homework Helper

    I'm picturing the two q's side by side with R straight down from the middle of them.
    Brilliant idea to find the force on R and use the 3rd law to get the answer you want!!!
    Most impressive.

    I'm not comfortable with the dipole formula, which gives no direction. If you know how to use it, it will be a shortcut.

    I would just sketch the E arrows on R due to each of the q's.
    Calculate the force using F = kq1*q2/r^2 for each.

    The direction of their total is obvious from symmetry. You'll have to do a sine to find the component of the F you want. It says p is much greater than s, and in that case tan(A) is approximately equal to sin(A), which will simplify things!
     
  4. Feb 5, 2009 #3
    Good idea to calculate the force instead of using the dipole formula. I think the directions will work out better that way.

    Due to symmetry, force on R is either towards the left or right, depending which way your dipole is oriented. The up and down parts cancel, right?

    And therefore, force on the dipole is opposite that.

    Hmm, just thought of another problem. The problem doesn't say if R is positively or negatively charged.
     
  5. Feb 5, 2009 #4
    Also, can you do the problem without using tan? Can I just use sin even if it will be more complicated?
     
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