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Energy of a dipole

  1. Dec 30, 2013 #1
    I have a dipole such as:

    [itex]\rho(\vec{r}) = q \delta(\vec{r} - \vec{a}) - q \delta(\vec{r} + \vec{a})[/itex]

    with [itex]\vec{a} = a \vec{e}_x[/itex].

    I have to show that the energy in a constant external field [itex]\vec{E}[/itex] is:

    [itex]V = - 2 q \vec{a} \vec{E}[/itex]

    My calculations so far:

    With the formula: [itex]\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \! \rho(\vec{r}') \frac{1}{|\vec{r} - \vec{r}'|} \, d^3r' [/itex]

    I have calculated for the potential:

    [itex]\phi(\vec{r}) = \frac{q}{4 \pi \epsilon_0} (\frac{1}{|\vec{r} - \vec{a}|} - \frac{1}{|\vec{r} + \vec{a}|})[/itex]

    The energy [itex]V(\vec{r}) = \phi(\vec{r}) q[/itex]

    Thus I get for the energy:

    [itex]V(\vec{r}) = \frac{q^2}{4 \pi \epsilon_0} (\frac{1}{|\vec{r} - \vec{a}|} - \frac{1}{|\vec{r} + \vec{a}|})[/itex]

    However I don't understand why that is equal to [itex]-2q \vec{a} \vec{E}[/itex].

    Any ideas?
     
  2. jcsd
  3. Dec 30, 2013 #2

    rude man

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    The energy referred to is the difference in energy between the dipole's x and y orientation. Depending on the direction of the p and E vectors, this difference can be positive or negative. So you must be careful with your signs.

    So look up the formula for the energy of a dipole in an E field. This expression is a dot product so be aware of your signs. Among other things you have to determine the sign (direction) of your dipole as given in the expression for ρ. Hint: ρ is a line charge density and the units of δ(x) are always 1/x.

    Alternatively, you can use the formula for torque on a dipole in an E field, which is also a vector expression, then compute the work needed to change the dipole unit vector from the x to the y direction. This energy can again be positive or negative so again be careful with your signs.

    P.S. Unless another useless change has occurred in notation since my day, your answer should indicate a dot product. There is no such thing as AB where A and B are vectors. It's either A dot B or A cross B.

    But times change, there is no such thing as relativistic mass anymore, and E = mc^2 is no longer correct if m moves. What a shame ...
     
  4. Dec 31, 2013 #3
    Hello rude man!

    We've learned the convention that you usually leave out the dot product sign if it's clear from context how it's meant to be. Just like you'd write 2x instead of "2 times x", we write [itex]\vec{A}\vec{B}[/itex] instead of [itex]\vec{A} \cdot \vec{B}[/itex] :smile:

    I don't think the + or - signs are the problems here. The charge density is given as it is, so the + or - signs automatically follow correctly from integration. You can't do much wrong integrating delta functions I think. The potential should be correct. My problem is to show why [itex]W = - 2 q \vec{a} \cdot \vec{E}[/itex] with [itex]\vec{E}[/itex] being an external field.

    The left hand side should be "potential times q". Where do you get the right hand side from though?
     
  5. Dec 31, 2013 #4

    rude man

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    I guess I don't think of potentials in this case. I guess you can but I don't. I seem to remember a paper where the approach is in that direction. Do a google and I think you can find it. It was by a Princeton co-author, a pdf file.

    Reason I don't understand the idea of potentials is that the dipole already has its own energy: -kq^2/(2a)^2, without the introduction of an external E field. But this term with k in it does not appear in the answer. You'd think adding the E field would be a sort of superposition of the dipole's intrinsic potential energy plus a term with E in it. But it isn't. So I view it as the energy associated with rotation, which does provide the right answser.

    If you do decide to go it my way:
    Did you find the equation for the energy of a dipole in an E field? Or at least the expression for torque on a dipole in an E field? That is how I would proceed.

    BTW I still think you need to know the direction of the dipole vector in relation to the direction of the E vector. If you got it backwards you'd get the wrong sign in your answer.

    Happy figuring!
     
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