What is the force on an elementary dipole from a point charge in the same plane?

[andre220]

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]

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 $\vec p$.

 

[Goddar]

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

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 p⋅r = pr cosθ, which should simplify your calculations...

 

[Goddar]

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

 

[andre220]

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.

 

[Goddar]

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...