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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.

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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. –  yohBS Feb 20 '12 at 6:59

1 Answer 1

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

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Your answer is right but in practice for his problem I think the solution in the comment is faster. In your general solution you need to have all derivatives of spherical coordinates with respect to cartesian ones and reciprocally, which is not easy - or at least fast! –  MrBrody May 24 '12 at 3:07

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