Have a point charge and a perfect dipole $\vec{p}$ a distance $r$ away. Angle between $\vec{p}$ and $\hat{r}$ is $\theta$. Want to find force on dipole.
I'm having more than a little difficulty identifying where I'm going wrong. If I do this problem in cartesian coordinates, I get the right answer, so apparently I am not understanding something about spherical coordinates.
We have $F = q\Delta E$ for dipoles in a nonuniform electric field. If $d$ in dipole is small, then I can use
$$\Delta E \approx \nabla E \cdot \Delta\vec{r}$$
Below I derive the expression in spherical coordinates.
So, first of all,
$$E = \frac{q}{4 \pi \epsilon_0 r^2} \hat{r}$$
So
$$E_r = \frac{q}{4 \pi \epsilon_0 r^2}$$
and
$$\Delta E_r = \nabla E_r \cdot \Delta \vec{r}$$
where $\Delta \vec{r} = \bigl(\Delta r, r\Delta \theta, r\sin\theta\Delta \phi \bigr)$.
$$\nabla E_r = \biggl(\frac{-2q}{4 \pi \epsilon_0 r^3},0,0\biggr)$$
Therefore,
$$q\Delta E_r = \frac{-2qp\cos\theta}{4 \pi \epsilon_0 r^3}$$
and
$$\Delta E_{\theta} = \Delta E_{\phi} = 0$$
as $E_{\theta} = E_{\phi} = 0$.
So
$$F = q\Delta E_r = \frac{-2qp\cos\theta}{4 \pi \epsilon_0 r^3} \hat{r}$$
But should be
$$F = \frac{-2qp\cos\theta}{4 \pi \epsilon_0 r^3} \hat{r} - \frac{qp\sin\theta}{4 \pi \epsilon_0 r^3} \hat{\theta}$$
So $\Delta E_{\theta}$ must be nonzero but I don't see how.