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Electric Field on a Water Molecule

  1. Jan 22, 2007 #1
    1. The problem statement, all variables and given/known data

    A water molecule is a permanent dipole with a dipole moment p = qs. There is a water molecule in the air a very short distance x from the midpoint of a long rod of length L and uniform charge +Q. The axis of the rod is perpendicular to the rod. Note that s << x << L.

    a) Find the the force of the rod on the molecule in terms of Q, L, x, p.

    b) If the electric field on the molecule when it is 1 cm from the rod is 10^6 N/C and p = 6*10^-30 and the mass of the molecule is m. What is the acceleration of the molecule?

    2. Relevant equations

    F = ma For a rod: F = k(Q/r*sqrt(r^2 + (L/2)^2)), k = 9*10^9

    F = qE
    3. The attempt at a solution

    I have done a and get:

    Force of rod on molecule: F = 2kQp/Lx^2
    (toward the rod and perpendicular to it)

    My problem is that I'm not sure how to get b. I don't know how to eliminate using given data, Q and L. Please help. Thank you.
  2. jcsd
  3. Jan 22, 2007 #2


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    Staff: Mentor

    I think you meant to type "the axis of the dipole is perpendicular to the rod", right?

    In this type of problem, the key is to realize that the electric field is diverging as it moves away from the rod, so that one charge sees a higher field than the other. That is what creates a net force on the dipole.

    Start assuming a positive charge on the rod, and write the equation for the Electric field as a function of radius away from the rod. Then assume that the dipole is set up initially with the - end toward the rod and the + end away. Now, what is the net force on the dipole? What does the dipole molecule do? Which way will it move?

    Now as a quiz question, what is different if the dipole is started reversed, with the + end in toward the rod and the - end away? (Hint -- is there a torque on the molecule if it starts to rotate?)
    Last edited: Jan 22, 2007
  4. Jan 22, 2007 #3

    But if you look at my ATTEMPT FOR A SOLUTION you will see this is precisely what I have already solved. I am still not sure about part b. Please read what I said about part b.

    With respect to the quiz, I can solve another problem as soon as I solve the first. Thanks anyways.
  5. Jan 22, 2007 #4


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    I don't understand your answer to a) ... what does the length L of the rod have to do with it? For a long charged rod, the Electric field has nothing to do with L, just with the linear charge density and the radius away from the rod.

    And for b), it look like they are assuming that you can calculate the force from the dipole moment p, not knowing the explicit values of the charges and separation. Does you textbook derive that relation? If not, you can derive it yourself by assuming some separation of the charges, and then making the simplifying assumption in the problem of s << x << L.
  6. Jan 22, 2007 #5
    Really? Well the formula I gave in the relevant equations section does seem to imply that L has something to do with E. It is also the equation from the book.

    Also, if you look at my answer you will see Q/L which is linear density?

    I don't seem to get what u r saying with respect to b. Not sure where should I start. Thanks
  7. Jan 22, 2007 #6


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    Staff: Mentor

    I googled force on electric dipole and got lots of good hits (in case your text isn't explaining it very well). Look at the end of this one to see if it helps:


    And on the use of L and Q to get a linear charge density, I suppose that's okay, as long as your equation in the OP is correct. I'm more used to using [tex]\rho_L[/tex] for the linear charge density, but I see now that the question did not give that to you except as the total L and Q.
  8. Jan 22, 2007 #7
    good website, and I see where u were going with the quiz. In one of its examples, the electric field due to a disk with uniform charge is derived. Unfortunately, another thing that I have been trying to find, is the derivation of the electric field due to a rectangular sheet. This I haven't found. Could you help me out with that?

    But I'm still struggling with part b... Thanks.
  9. Jan 22, 2007 #8


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    Staff: Mentor

    I posted a hint in your rectangular charge sheet thread. Sorry nobody got to it for a while. Keep in mind that a lot of homework helpers and Mentors may not end up checking the PF much over the weekend. I do it when I can, but it was a busy weekend for me.

    Also, here's a tip about threads. At least for me, when I scan a forum looking for threads to open to see if I can be of help, I look mainly for 0-reply threads, since I know the original poster (OP) hasn't gotten any help yet. When I'm in a hurry, I may not even skim the thread topics or poster names -- I'll just look for 0-reply threads and then pick some of those where the thread title sounds like I might be able to help. So when you reply to your own thread a couple times to try to BUMP it back up in the thread queue, you may actually be hurting your chances of getting help. To counteract this somewhat, if I'm one step down from in a big hurry, I'll also see if there are any threads with 1 reply, and the OP and the last reply are the same person. But when a thread has multiple replies and the last reply is the OP, then it's a lot less likely I'll stop in (because I'll assume that you're getting help and replying to that help).

    Just a tip. With the PF being all-volunteer, sometimes help comes faster than other times.
  10. Jan 22, 2007 #9


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    Staff: Mentor

    Oh, and on your part b), all you need is to find (or derive) the equation for the force on a small electric dipole in a diverging electric field, in terms of the dipole moment p (instead of the q and s values themselves). I would think that if you re-googled the terms that I mentioned, another hit on that list should be able to help you. Or maybe add in the term moment to narrow the search a bit. Sorry that I'm of no more help at the moment. Are you sure this isn't covered in your text? It's in section 4.3 of my old electrostatics book by Plonus.
  11. Jan 23, 2007 #10
    I looked for it, but the only useful thing I found, I think, was the torque on the dipole in terms of E and p. But I don't believe there to be a torque on the molecule since the electric field due to the rod is already aligned with the molecule's dipole moment. Could you please show me the derivation?
  12. Jan 23, 2007 #11


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    I googled electric dipole force in non-uniform electric field, and got lots of good hits. Here is one of the first ones, which guides you through the derivation. If this hit doesn't help you, try googling with my terms and check out the other hits.

    http://schneidm.grinnell.edu/d2pdf/9.pdf [Broken]
    Last edited by a moderator: May 2, 2017
  13. Jan 23, 2007 #12


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    This is correct.

    You do not need to eliminate Q and L separately. Notice that the expressions for the field and the force only involve the ratio Q/L. You can find the value of Q/L from the given field at x=1cm. If you plug in this value, and the value of p (given), you can find a numerical value for F (force on dipole).
  14. Jan 23, 2007 #13
    But doesnt the field not include p in its expression? If you take my answer and divide by q to get the field, then doesn't p disappear to be replaced with s? If so, how is it that I get Q/L? Thanks
  15. Jan 24, 2007 #14
    Oh, is it that perhaps I have a good idea? Idea: since water is a good conductor (almost as good as a metal) then the net electric field, say, at the center of the water molecule, must be zero. That is, the sum of the electric fields produced by the rod and the polarized charges at the opposite edges of the molecule, at the center of the molecule must be zero. This is a property of metals and conductors. If that is true, then check my work. An expression for the force is derived that that doesn't have any unknowns.

    1) Electric field at the center of the molecule:

    ksQ/x^2L - k2q/(s/2)^2 = 0

    q = s^3Q/8x^2L

    2)Force on the molecule:

    F = qE = s^3QE/8x^2L

    kpQ/x^2L = s^3QE/8x^2L

    s = (8kp/E)^(1/3)

    3) Dipole Moment:

    p = qs ==> q = p/s = p(E/8kp)^(1/3)

    4) Finally, the expression for the force:

    F = qE = pE(E/8kp)^(1/3)

    F = 1/2 * (kp^2 E^4)^(1/3)

    Please check my work. Note that this expression is only the result of the assumption made that the water molecule has similar conducting properties to that of a metal. (net electric field inside the molecule is zero). Is there an easier way to get around to the same answer? Something tells me there is...
    Last edited: Jan 24, 2007
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