Force on an elementary dipole

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  • #1
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Homework Statement


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}##.

Homework Equations


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

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?
 

Answers and Replies

  • #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##.
 
  • #3
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$$F = -\frac{d\Phi(\vec{r})}{dr}$$
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...
 
  • #4
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Sorry, didn't see Orodruin's post... I'll leave it to him from now on.
 
  • #5
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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.
 
  • #6
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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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