# Force on an elementary dipole

## 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?

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##.

$$F = -\frac{d\Phi(\vec{r})}{dr}$$

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...

Sorry, didn't see Orodruin's post... I'll leave it to him from now on.

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.

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...