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Electric field and orientation

  1. Jan 27, 2014 #1
    Suppose that H2O molecule (micro = 1.85 D) approaches an anion (such as Cl-) in the
    gas phase (situation typical of the ion-induced nucleation of atmospheric water vapor).
    (a) What is the favorable orientation of a molecule? (b) Calculate the electric field (in
    V/m) experienced by the anion when water dipole is 1.0 nm from the ion.

    2. Relevant equations
    I don't know if it is relevant but I think this must be used maybe E=F/q or E=K * Q/d^2 where K is a constant of 9.0x10^9 N * m^2/C^2 and Q is electric force and d is distance between the two objects

    3. The attempt at a solution
    Okay so I don't know if this is right but this my guess E= (9.0x10^9 * 1.85)/(1.0x10^-9)^2
    = 1.67x10^28 N*D/C^2 I know this is not the right units and that to get voltage i need to have N*m/C that then equals joules/C that =voltage. I am just very confused and any help would be great. I am not exactly sure what it means by orientation like if is talking about the geometry of the molecule or if it should be drawn out with the hydrogen bonds on the water molecule facing towards the anion.
  2. jcsd
  3. Jan 27, 2014 #2

    Simon Bridge

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    You should be reasoning it out rather than guessing.

    You need to show the favorable orientation ... this just requires a qualitative understanding of electrostatics.
    The water molecule is being modeled as a dipole - you should have something about this in your notes.
    The Cl- ion, by comparison, is just a normal negative charge.

    So which way will the water molecule want to orient wrt the ion?
    i.e., which charges attract and which repel?

    The second part requires the electric field of a dipole - which will be in your notes.
    It looks like you are given the dipole moment if by "micro" you mean the Greek letter "mu".
  4. Jan 28, 2014 #3
    Actually we haven't covered any of this in class we have looked at computer simulations of water in class and thats about it so far so it is me just guessing cause we have not been given any formulas. The formula i have in my thing is from me looking it up online since there is also no textbook for this class.
  5. Jan 28, 2014 #4
    As well I think i solved it where i turned the Debye unit into C*m so this is what i got
    (9.0x10^9)(6.17x10^-36)/(1.0x10^9)^2=0.05553N*C which equals 0.05553V/m.
  6. Jan 28, 2014 #5

    Simon Bridge

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  7. Jan 29, 2014 #6
    welll the orientation is going to be the water molecules surrounding the anion where the positive hydrogen's are going to be towards the negative anion charge. The equation looked like it fit the question lol.
  8. Jan 29, 2014 #7

    Simon Bridge

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    Since it's a dipole, the water molecule is being modeled as a kinda dumb-bell with a + charge on one end and a - charge on the other end.

    What you've just said is that the + end will point towards the anion - good.
    You are asked to calculate the electric field.

    I am concerned that you don't understand the relations you have been using.

    how does the equation account for the fact there are two charges in the dipole?
    if the orientation were not end-on like that, would that make a difference to the equation?

    finally: electric field is a vector. You were not asked for just the magnitude so you should probably include a statement about the direction as well.

    Did you have a look at the link I gave you?
  9. Jan 29, 2014 #8
    well I assumed since hydrogen is a +1 charge and the anion like chlorine is a -1 charge that the charges would cancel each other out. I thought that the charges had to be accounted for if one had like a +2 and one had a -1. I would assume if the orientation was not end on end than yes it would change the equation because there would not be the interaction of the anion and cation. thus not creating a bond. I looked at the link I would like it if they explained what everything is in the equations. Cause I got slightly confused with it and thank you for your help. I wish this was covered in class.
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