# 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.

• The general formula for the force is, as you correctly stated $F=q\Delta E$, which is conveniently written in a geometrical form as the dot product $$\vec F=\vec\nabla\vec E \cdot q\Delta\vec r =\vec \nabla\vec E \cdot\vec P$$ Note that $\vec\nabla\vec E$ is a matrix. However, when you work in any other coordinate system, the gradient $\vec \nabla$ is no longer the simple expressions that you are used to. You can derive the formula for it by differentiating the expression $$\vec E=E_r\hat r+E_\phi \hat \phi+E_\theta\hat\theta$$ and remembering that the unit vectors are also space-dependent. Feb 20, 2012 at 6:59

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 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. $$F^{\,i} = T{\,}^{ji}_{t}(y) \delta^t_j \quad \text{(wrong)}$$