Force from point charge on perfect dipole 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.
 A: The force applied to a point dipole with dipole momentum $\vec{p}$ is
$$
\vec{F} = (\vec{p} \cdot \vec\nabla) \vec{E}
$$
In Cartesian coordinates that is
$$
F_i = \sum_j p_j \frac{\partial}{\partial x_j} E_i
$$
But in spherical coordinates it is not the same.
There is no field components along $\vec{\theta}$, but there is a gradient of field components along this direction since the direction of the vector changes.
In order to convert this expression to spherical coordinates one should to use tensor analysis.
In all following expressions the summation over repeating indices is assumed.
$$
T^{\;ji}_t = p^j \frac{\partial}{\partial x^t} E^i
$$
$$
F^{\;i} = T^{\;ji}_t \delta^t_j
$$
Let Cartesian coordinates be $x^1, x^2, x^3$ and spherical coordinates be $y^1, y^2, y^3$, then
$$
T{\,}'^{j'i'}_{t'}(y) =
\frac{\partial y^{j'}}{\partial x^j}
\frac{\partial y^{i'}}{\partial x^i}
\frac{\partial x^t}{\partial y^{t'}}
T^{\;ji}_{t}\bigl(x(y)\bigr)
$$
One should calculate the force in spherical coordinates as
$$
F^{\,i} = T{\,}'^{ji}_{t}(y) \delta^t_j \quad \text{(correct)}
$$
while you have used the tensor without prime, i.e.
$$
F^{\,i} = T{\,}^{ji}_{t}(y) \delta^t_j \quad \text{(wrong)}
$$
