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Force on an electric dipole

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data

    HnUp0.jpg

    Two electric dipoles, oriented as shown in the figure, are separated by a distance r.

    What is the force on p1 due to p2

    2. Relevant equations

    F = (p dot ∇)E

    3. The attempt at a solution

    I'm confused on the p dot ∇ part.

    How is the divergence of p not just zero?

    It seems p1 would just be some constant in the x hat direction.
     
  2. jcsd
  3. Dec 11, 2012 #2

    diazona

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    Think about that carefully. [itex]\mathbf{p}\cdot\nabla[/itex] is not the divergence of [itex]\mathbf{p}[/itex]!
     
  4. Dec 12, 2012 #3
    I'm not sure I understand that. ∇⋅p would be divergence? And dot products are commutative.
     
  5. Dec 12, 2012 #4

    diazona

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    This isn't really a dot product, though, it's a dot "application." That is, it still means that [itex]\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z[/itex], but you can't assume that e.g. [itex]a_x b_x[/itex] means "[itex]a_x[/itex] multiplied by [itex]b_x[/itex]," because you are dealing with things that don't get multiplied. Think about it: does [itex]x \frac{\partial}{\partial x}[/itex] mean [itex]x[/itex] multiplied by [itex]\frac{\partial}{\partial x}[/itex]? Does [itex]\frac{\partial}{\partial x} x[/itex] mean [itex]\frac{\partial}{\partial x}[/itex] multiplied by [itex]x[/itex]?
     
  6. Dec 12, 2012 #5
    Oh ok so if the vector is in front of del i'm basically taking the components and multiplying them by differential operators, where if the vector was to the right of del i would be taking derivatives of the components?
     
  7. Dec 12, 2012 #6

    diazona

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    Yep, that's the idea.

    A good way to think about it is that the components of [itex]\mathbf{p}[/itex] are multiplicative operators. Just like the differential operator [itex]\frac{\partial}{\partial x}[/itex] takes a function [itex]f(x)[/itex] and turns it into [itex]f'(x)[/itex], a multiplicative operator [itex]x[/itex] takes a function [itex]f(x)[/itex] and turns it into [itex]xf(x)[/itex]. The composition of two operators is also an operator, so [itex]p_x \frac{\partial}{\partial x}[/itex] is an operator that means "take the derivative and then multiply by a number."
     
  8. Dec 12, 2012 #7
    Thanks, really helped me.
     
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