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Force on an elementary dipole

  1. Feb 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that the force on an elementary dipole of moment ##\mathbf{p}##, distance ##\mathbf{r}## from a point charge ##q## has components
    $$\begin{eqnarray}
    F_r &=& -\frac{qp\cos{\theta}}{2\pi\epsilon_0 r^3}\\
    F_\theta &=& -\frac{qp\sin{\theta}}{4\pi\epsilon_0 r^3}
    \end{eqnarray}$$
    along and perpindicular to ##\mathbf{r}## in the plane of ##\mathbf{p}## and ##\mathbf{r}##, where ##\theta## is the angle which ##\mathbf{p}## makes with ##\mathbf{r}##.

    2. Relevant equations
    $$\Phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\frac{\vec{p}\cdot\vec{r}}{r^3}$$
    $$F = -\frac{d\Phi(\vec{r})}{dr}$$
    $$\vec{p} = Q\vec{r}$$

    3. The attempt at a solution
    Frankly, I do not know how to start this one. I need to find the force on the charge from the dipole. To do so, I take the derivative of ##\Phi## and find the force using Coulomb's law equation. And them decompose it in terms of ##r## and ##\theta##. Is this the right direction?
     
  2. jcsd
  3. Feb 2, 2015 #2

    Orodruin

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    Note that the force is the gradient of the potential. What you have written down in the relevant equations is simply the r-component.

    Also, please show us what you get when you try to do what you said.

    Note: ##Q\vec r## is the dipole moment of a point charge ##Q## in ##\vec r## with respect to the origin. This is not the situation in your problem, which has a fundamental dipole ##\vec p##.
     
  4. Feb 2, 2015 #3
    Hi. The above equation is not right and you can tell because force is a vector and your expression gives you a scalar.
    In 3 dimensions: F = –∇Φ, which is a gradient and you'll have to look up how to take it in spherical coordinates because the expression is far from obvious.
    Lastly, without loss of generality you can take p to be along the z-axis so that pr = pr cosθ, which should simplify your calculations...
     
  5. Feb 2, 2015 #4
    Sorry, didn't see Orodruin's post... I'll leave it to him from now on.
     
  6. Feb 4, 2015 #5
    Yes, thank you. It was quite simple once you pointed out the fact that ##F## should be a vector. And that ##p\cdot r = p r \cos{\theta}##. Thank you for your help.
     
  7. Feb 4, 2015 #6
    No problem. I just have to make it clear that although you have in general p⋅r = pr cosδ, where δ is the angle between p and r, you can take δ = θ (the polar angle of your spherical coordinates) only if you have freedom in choosing the direction of p, which is the case here. It's a detail but always important to understand clearly...
     
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